I need help finding the force of friction please.

Question

"A shopper pushes a 28N cart a distance of 53 m along a level grocery store surface. The force he exerts is 35 N at an angle of 42 below the horizontal. The coefficient of friction is .15. Find the work he does on the shopping cart. Find the work done by the force of friction on the shopping cart." I have found that the work is 1378.5, but I don't know what equation to use to find the work done by the force of friction.

Answer 1

Let

F = 28 N be the force,

s = 53 m be the distance,

α = 42° be the angle of F below the horizontal,

f = 0.15 be the coefficient of friction,

g = 9.81 m/s² be gravitational acceleration,

m (unknown) be the mass of the cart.

The person's force F can be decomposed into two orthogonal forces.

- A horizontal forward force of magnitude Fcos(α).

- A downward force of magnitude Fsin(α).

The horizontal force performs work

Fcos(α)s = 28×cos(42°)×53 = 1102.8 J

Friction is caused by downward forces, of which there are two:

- the weight of the card mg,

- the person's downward force Fsin(α).

The force of friction is then the total vertical force multiplied by the coefficient of friction.

That is, the force of friction is (mg + Fsin(α))f.

Since the force of friction is directed against the direction of movement,

the work performed by friction is negative:

-(mg + Fsin(α))fs

The mass m of the cart is not given. If we assume that m = 0, then the work of friction is

-Fsin(α)fs = -28×sin(42°)×0.15×53 = -149 J