Question

A 50 kg crate slides down a 5.0 m loading ramp that is inclined at an angle of 25 to the horizontal. A worker pushes on the crate parallel to the surface of the ramp so that the crate slides down with a constant velocity. If the coefficient of kinetic friction between the crate and the ramp is 0.33, how much work is done by (a) the worker

Answer 1

Answer:

The magnitude of the work done by the worker is 303 J.

Explanation:

The work done by the worker can be found as follows:

$W = |F|\cdot |d| cos(\alpha)$                              

Where:

F: is the force applied by the worker

d: is the displacement = 5.0 m

α: is the angle between the force applied and the displacement = 180°

We need to find the force applied by the worker:

$\Sigma F = ma$

Taking as positive the movement direction of the crate we have:

$-F - F_{\mu} + P_{x} = 0$  

Where:              

m: is the crate's mass

a: is the acceleration = 0 (It is moving at constant speed)

F: is the force applied by the worker

Pₓ: is the weight in the horizontal direction    

$F_{\mu}$: is the frictional force  

Hence, the force applied by the worker is:            

$F = P_{x} - F_{\mu} = mgsin(\theta) - \mu mgcos(\theta)$

$F = 50 kg*9.81 m/s^{2}*(sin(25) - 0.33cos(25)) = 60.6 N$  

Then, the work done by the worker is:                        

$W = |F|\cdot |d| cos(\alpha) = 60.6 N*5.0 m*cos(180) = -303 J$                              

Therefore, the magnitude of the work done by the worker is 303 J.

I hope it helps you!                                            

Answer 2

The work is done by the worker will be 303 J. Work done is described as the multiplication of applied force and the amount of displacement.

What is work done?

Work done is defined as the product of applied force and the distance through which the body is displaced on which the force is applied.

Work may be zero, positive and negative.it depends on the direction of the body displaced. if the body is displaced in the same direction of the force it will be positive.

The given data in the problem is;

F is the force applied by the worker

d is the displacement = 5.0 m

α is the angle between the force applied and the displacement = 180°

m is the crate's mass=  50 kg

a is the acceleration = 0

Pₓ: is the weight in the horizontal direction    

The net force on the crate is found as;

$\rm F_{net}= P_X - F_{\mu} \\\\ \rm F_{net}= mg sin \theta - \mu mg cos \theta \\\\ \rm F_{net}=50 \times \times 9.81 (sin 25^0 -0.33 cos(25) \\\\ \rm F_{net}= 60.6 N$

The work done  by the worker will be;

$\rm W= Fd cos \alpha \\\\ \rm W= 60.6 \times 5.0 cos 180^0 \\\\\rm W=-303 \ J$

Hence the work is done by the worker will be 303 J.

To learn more about the work done refer to the link ;

https://brainly.com/question/3902440