a playful kid pushes a grocery cart on a frictionless floor with a constant force of 10 n. the grocery nah cart has a mass of 10 kg and it is at rest before the kid pushes it. if her mother yelled at her after pushing the grocery cart at displacement of 6.7 meters, what is the final velocity of the grocery cart during this time?
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Answers
(a) The 35N force applied by the shopper makes a 25
0
angle with the displacement of the cart (horizontal). The work done on the cart by the shopper is then
W
shopper
=(Fcosθ)Δx=(35.0N)(50.0m)cos25.0
0
=1.59×10
3
J
(b) The force exerted by the shopper is now completely horizontal and will be equal to the friction force, since the cart stays at a constant velocity. In part (a), the shopper’s force had a downward vertical component, increasing the normal force on the cart, and thereby the friction force. Because there is no vertical component here, the friction force will be less, and the the force is smaller than before.
(c) Since the horizontal component of the force is less in part (b), the work performed by the shopper on the cart over the same 50.0m distance is the same as in part (b).
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(a) The displacement of the cart creates a 25-degree angle with the 35N force that the shopper applies (horizontally).
therefore,
therefore,
therefore,
(b) Since the cart continues to move at a steady speed, the force applied by the shopper is now entirely horizontal and will be equal to the friction force. The shopper's force in portion (a) had a downward vertical component, which raised the normal force acting on the cart and, as a result, the friction force. The force is less than before since there isn't a vertical component in this situation, which reduces the friction force.
(c) The work done by the shopper on portion (b) is less because the horizontal component of the force is smaller.
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