1) When all forces are balanced the resulting acceleration is a) 0 m s-2 b) in the absence of any external forces a body will remain at rest or in constant velocity in a straight line c) force, Newtons d) mgh = 1/2 mv2 , the m cancels out on both sides leaving gh = 1/2 v2 2) What is Newton's First Law a) The air resistance increases to much more than the weight b) in the absence of any external forces a body will remain at rest or in constant velocity in a straight line c) F = 288 N d) 3) What is Newton's Second Law a) The acceleration of an object is proportional to the force applied and inversely proportional to its mass b) Friction is a contact force that opposes the direction of motion c) W = mg sin(15) d) They have reached terminal velocity and are no longer accelerating 4) What equation is associated with Newton's Second Law a) 40 m b) F = ma c) The acceleration of an object is proportional to the force applied and inversely proportional to its mass d) 5) What is Newton's Third Law a) The air resistance increases to much more than the weight b) For every action there is an equal and opposite reaction c) W = mg cos(15) d) Change speed, change direction and change shape of an object 6) What are the three effects a force can have? a) Ew = 15 cos(25) x 20 = 270 N b) F = ma c) F = 288 N d) Change speed, change direction and change shape of an object 7) What is the value of gravitational field strenght near the surface of Earth? a) The air resistance increases to much more than the weight b) 19600 J c) W = mg sin(15) d) 9.8 m s-2 8) Weight is a _____________ measured in ___________________. a) W = mg b) Right angled triangle with Weight as the hypotenuse. c) force, Newtons d) F = 5.0 N 9) The formula connecting weight and mass is: a) F = 288 N b) W = mg sin(15) c) They are in constant freefall which is due to the curvature of the path of the ISS around the Earth d) W = mg 10) What effect does friction have on a moving object a) Friction is a contact force that opposes the direction of motion b) Ew = 15 cos(25) x 20 = 270 N c) Ew = Fd = 40 x 10 = 400 J d) They are in constant freefall which is due to the curvature of the path of the ISS around the Earth 11) Why is a parachute no use when landing on Mars? a) E = Pt = mgh => t = mgh/P, t = 2.6 s b) Ew = Fd = 60cos(30) x 10 = 520 J c) There is no real atmosphere to provide a counter force to the gravitational pull. d) Friction is a contact force that opposes the direction of motion 12) Write an equation for the component of the weight force parallel to a uniform slope. The slope has an angle of 15 degrees above horizontal a) The acceleration of an object is proportional to the force applied and inversely proportional to its mass b) W = mg sin(15) c) Change speed, change direction and change shape of an object d) mgh = 1/2 mv2 , the m cancels out on both sides leaving gh = 1/2 v2 13) Write an equation for the component of the weight force perpendicular to a uniform slope. The slope has an angle of 15 degrees above horizontal a) False - the exhaust gases push against the rocket, not space. b) Gravitational force and air resistance c) W = mg cos(15) d) F = 288 N 14) A block on a ramp is at rest. It has a mass of 1.5 kg and the ramp is at 20° above horizontal. What is the friction force? a) Change speed, change direction and change shape of an object b) The weight force remains the same c) F = 5.0 N d) The air resistance increases to much more than the weight 15) What is the size of the normal reaction force to a book of 750 g mass resting on a table a) 19600 J b) mgh = 1/2 mv2 , the m cancels out on both sides leaving gh = 1/2 v2 c) F = 7.4 N d) Ew = Fd or W = Fd 16) A sky diver with a mass of 60.0 kg is accelerating at 5.00 m s^-2. What is the force of friction from air resistance at that instant? a) E = Pt = mgh => t = mgh/P, t = 2.6 s b) F = 288 N c) Change speed, change direction and change shape of an object d) F = 5.0 N 17) A sky diver with a mass of 60.0 kg is at terminal velocity. What is the force of friction from air resistance at that instant? a) E = Pt = mgh => t = mgh/P, t = 2.6 s b) 27.3 m/s c) They have reached terminal velocity and are no longer accelerating d) F = 588 N 18) What is the initial vertical air resistance when a sky diver steps out of a plane? a) F = 588 N b) E = Pt = mgh => t = mgh/P, t = 2.6 s c) The engine exerting and equal and opposite force on the expelled xenon ion d) F = 0 N, there is no vertical air resistance at the start of freefall 19) What forces are involved when an object is in freefall? a) F = 0 N, there is no vertical air resistance at the start of freefall b) in the absence of any external forces a body will remain at rest or in constant velocity in a straight line c) 9.8 m s-2 d) Gravitational force and air resistance 20) What happens when a sky diver's air resistance is equivalent to their weight when falling? a) There is no real atmosphere to provide a counter force to the gravitational pull. b) They have reached terminal velocity and are no longer accelerating c) Ew = Fd = 60cos(30) x 10 = 520 J d) Ew = 15 cos(25) x 20 = 270 N 21) How does the weight force change when a person is sky diving? a) b) The weight force remains the same c) F = 5.0 N d) v2 = 2gh => v = 25 m/s 22) What happens to the friction force when the parachute is opened? a) Change speed, change direction and change shape of an object b) 9.8 m s-2 c) The air resistance increases to much more than the weight d) 23) In space rockets can accelerate because the exhaust gases push against space. True or False a) False - the exhaust gases push against the rocket, not space. b) v2 = 2gh => v = 25 m/s c) W = mg cos(15) d) Change speed, change direction and change shape of an object 24) In an ion drive positive ions of xenon are repelled by an electric field out from the engine. What is the Newton pair in this process? a) They are in constant freefall which is due to the curvature of the path of the ISS around the Earth b) The weight force remains the same c) W = mg sin(15) d) The engine exerting and equal and opposite force on the expelled xenon ion 25) Why do astronauts in the International Space Station appear weightless although they are clearly still in the gravitational field of the Earth? a) They are in constant freefall which is due to the curvature of the path of the ISS around the Earth b) The engine exerting and equal and opposite force on the expelled xenon ion c) Friction is a contact force that opposes the direction of motion d) 0 m s-2 26) How much work is done on a box horizontally moved across a 10 metre room? The force used to move the box was a constant 40 N horizontally. a) F = 5.0 N b) 0 m s-2 c) Ew = Fd = 40 x 10 = 400 J d) The engine exerting and equal and opposite force on the expelled xenon ion 27) How much work is done on a box moved horizontally across a 10 metre room by a force at 30 degrees above horizontal with a size of 60 N? a) False - the exhaust gases push against the rocket, not space. b) The engine exerting and equal and opposite force on the expelled xenon ion c) Ew = Fd = 60cos(30) x 10 = 520 J d) force, Newtons 28) 19600 J of energy was used to raise a 50 kg box. How high was it raised? a) W = mg b) F = 5.0 N c) 40 m d) 27.3 m/s 29) A 50 kg box with a height of 40 m above ground has gained how much potential energy? a) F = 288 N b) v2 = 2gh => v = 25 m/s c) W = mg d) 19600 J 30) What speed would a box with mass 50 kg have if dropped from 40 m above ground, just before it lands on the ground? Air resistance may be ignored. a) False - the exhaust gases push against the rocket, not space. b) 27.3 m/s c) Ew = Fd = 40 x 10 = 400 J d) They are in constant freefall which is due to the curvature of the path of the ISS around the Earth 31) Draw a free body diagram for a book on an angled surface. The angle of the surface is 20 degrees above horizontal, and the book is stationary. a) 0 m s-2 b) 9.8 m s-2 c) W = mg cos(15) d) 32) What shape will the weight, normal reaction and friction force make when a book is stationary on a slope with an angle of 20 degrees? a) Right angled triangle with Weight as the hypotenuse. b) force, Newtons c) W = mg cos(15) d) F = 588 N 33) Write the equation connecting force, distance and energy Write the equation connecting force, distance and energy a) The acceleration of an object is proportional to the force applied and inversely proportional to its mass b) The engine exerting and equal and opposite force on the expelled xenon ion c) Ew = Fd or W = Fd d) F = ma 34) Why do we not need to know the mass of a pendulum bob when calculating how fast it is moving when released from a known height? a) Ew = Fd or W = Fd b) Ew = 15 cos(25) x 20 = 270 N c) mgh = 1/2 mv2 , the m cancels out on both sides leaving gh = 1/2 v2 d) E = Pt = mgh => t = mgh/P, t = 2.6 s 35) A box is moved horizontally 20 m by a pulling force of 15 N acting at 25° above horizontal. How much work is done on the box a) 9.8 m s-2 b) F = 0 N, there is no vertical air resistance at the start of freefall c) Ew = 15 cos(25) x 20 = 270 N d) F = 5.0 N 36) A box lifted 32 m above the surface is stationary. The cable holding it breaks. What is the speed of the box as it hits the ground? a) 0 m s-2 b) E = Pt = mgh => t = mgh/P, t = 2.6 s c) W = mg sin(15) d) v2 = 2gh => v = 25 m/s 37) A winch with a power of 150 W vertically raises a 9.7 kg block 4.1 m. How long did this take? a) E = Pt = mgh => t = mgh/P, t = 2.6 s b) For every action there is an equal and opposite reaction c) W = mg cos(15) d) They have reached terminal velocity and are no longer accelerating

Plane Higher Roulette - Forces

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