Asked by TT2014

A shopper pushes a grocery cart for a distance 18 m at constant speed on level ground, against a 25 N frictional force. He pushes in a direction 24° below the horizontal.
What is the work done on the cart by the shopper, in joules? Magnitude of the force, in newtons, that the shopper exerts on the cart? What is the total work done on the cart, in joules?

Answers

Answered by Henry
Fx - Fk = M*a
Fex*Cos24 - 25 = M*0 = 0
Fex*Cos24 = 25
Fex = 27.4 N. = Force exerted

Work = Fx*d = 27.4*Cos24 * 18 = 450.6 Joules

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