Question

A worker pushes a 1080 N crate with a horizontal force of 345 N a distance of 14 m. Assume the coefficient of kinetic frictin between the crate and the floor is 0.22. a) How much work is done by the worker on the crate

Answer 1

Answer:

The work done by the worker on the crate is 4830 J.

Explanation:

Given that,

Force = 1080 N

Horizontal force = 345 N

Distance d = 14 m

Coefficient of friction = 0.22

We need to calculate the work done

Using formula of work done

$W = Fd$

F = force

d = distance

Put the value into the formula

$W=345\times14$

$W=4830\ J$

Hence, The work done by the worker on the crate is 4830 J.

Answer 2

Answer:

(a) 1503.6 J

Explanation:

Weight of crate, mg = 1080 N

Applied force, F = 345 N

d = 14 m

μ = 0.22

(a)

Work done by the applied force, W1 = Appled force x distance

W1 = F x d = 345 x 14 = 4830 J

Work done by the friction force, W2 = Friction force x distance

W2 = μ x mg x d = 0.22 x 1080 x 14 = 3326.4 J

This work is negative as the direction of friction force and the distance is opposite to each other.

So, the work done by the worker on the crate is

W = W1 - W2 = 4830 - 3326.4 = 1503.6 J