Summary







































Walker5e CQ 5.09
If the force exerted by a horse on a cart is equal and opposite to the force exerted by the cart on the horse, how does the horse manage to move the cart?
a horse pulling a cart A. Gravity provides added force
B. Friction provides added force
C. Net force on cart is nonzero
D. Newton's 3rd Law doesn't apply here
Answer



















klm
A 85-kg hockey player on frictionless ice exerts a 48-N force on a 0.165-kg hockey puck, accelerating it at +291 m/s². What is the acceleration of the hockey player?
hockey player hitting a puck A. zero
B. −2.45×10−4 m/s²
C. −0.565 m/s²
D. −291 m/s²
Answer



















PoP5 4.16
You stand on the seat of a chair (yi = 0.50 m) and then hop off. Through what distance does the Earth (M = 6×1024 kg) rise up to meet you?
A. 10−23 m
B. 10−19 m
C. 10−15 m
D. 10−9 m
Answer



















Walker5e EYU 5.04
A force F pushes on three boxes that slide without friction on a horizontal surface, as indicated in the figure. The boxes are in contact, and slide as a group with an acceleration a. The magnitude of the force of contact between boxes 1 and 2 is _____ the magnitude of the force of contact between boxes 2 and 3.
force exerted on boxes of mass 1, 2, and 3 kg A. greater than
B. less than
C. equal to
Answer







































answer

C. Net force on cart is nonzero

is the best answer although one could also argue for

B. Friction provides added force

because the friction between the horse's hoofs and the ground is what propels the horse-cart combo.



















answer

C. −0.565 m/s²
Newton's third law states the force on the hockey player is equal and opposite to the force on the puck, so the player experiences a force of −48 N and an acceleration of (−48 N)/(85 kg) = −0.565 m/s²






















answer

A. 10−23 m
solution equation



















answer
force exerted on boxes of mass 1, 2, and 3 kg
A. greater than
Here it is best to think about Newton's second law as it applies to the box(es) to the right of each contact force. The force between boxes 1 and 2 must accelerate 5 kg but the force between boxes 2 and 3 need only accelerate 3 kg. Because the boxes are in contact we know their accelerations are the same, and we conclude F12 > F23. Practice applying Newton's second law for the case a = 4.0 m/s² and see if you get F = 24 N, F12 = 20 N, and F23 = 12 N.






















 
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