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A shopper in a supermarket pushes a cart with a force of 48.0 N directed at an angle of 25.0° below the horizontal. The force is just sufficient to balance various friction forces, so the cart moves at a constant speed.

a. Find the work done by the shopper as she moves down a 50.0 m length aisle.
b. What is the net work done on the cart?
c. The shopper goes down the next aisle, pushing horizontally and maintaining the same speed as before. If the work done by frictional forces doesn't change, would the shopper's applied force be larger, smaller , or the same? What about the work done on the cart by the shopper?

Respuesta :

Answer:

a) [tex]W=2175.14\ J[/tex]

b) [tex]W=2175.14\ J[/tex]

c) force required would be lesser, Work done will be constant

Explanation:

Given:

force, [tex]F=48\ N[/tex]

angle of force, [tex]\theta=-25^{\circ}[/tex]

(a)

displacement, [tex]s= 50\ m[/tex]

As the displacement is horizontal, we find the fomponent of the force in the direction of displacement.

[tex]F_H=F.cos\theta[/tex]

[tex]F_H=48\times cos(-25^{\circ})[/tex]

[tex]F_H=43.50\ N[/tex]

∴Work done:

[tex]W=F_H.s[/tex]

[tex]W=43.5\times 50[/tex]

[tex]W=2175.14\ J[/tex]

(b)

Net work done on the cart is same as the work done by the shopper.

[tex]W=F_H.s[/tex]

[tex]W=43.5\times 50[/tex]

[tex]W=2175.14\ J[/tex]

(c)

While pushing horizontally the force required would be lesser because it is all in the direction of displacement.

Work done will be constant because the force in the direction of a constant displacement would be the same.

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