Question

A worker pushes a wheelbarrow with a horizontal force of 50 N on level ground over a distance of 5.0 m. If a friction force of 43 N acts on the wheelbarrow in a direction opposite that of the worker, what work is done on the wheelbarrow by the worker?a) 250 J
b) 215 J
c) 35 J
d) 10 J
e) None of these answers is correct.

Answer 1

Answer:

Work done will be equal to 35 J

So option (c) will be correct answer

Explanation:

We have given force F = 50 N

Frictional force f = 43 N

So net force will be equal to $F_{net}=50-43=7N$

Distance covered on the ground d = 5 m

We have to find the work done

Work done is equal to $W=F_{net}\times d=7\times 5=35J$

So option (C) will be the correct answer.

Answer 2

The correct option is A 250 J.

Given, Horizontal  force is 50 N.

Frictional force is 43 N.

The distance between worker and pushing of wheelbarrow is 5 m.

We know that, Work is a physics term describing the amount of energy transferred when it is moved over a distance by an external force.

So Work = Force . Displacement

 work = F D cos $\Theta$......(eq 1)

Here $\Theta$ is 0 .

So, $work = 50 \times 5 \times 1$ ( cos 0 = 1 )

Work = 250 J

Hence 250 J of work is done on the wheelbarrow by the worker.

For more details on the work follow the link:

https://brainly.com/question/13594876