Why Is My Calculation of Work for Sliding a Crate Up a Ramp Incorrect?

[Xamfy19]

I have question regarding work and I was stuck with the wrong solutions.

A guy slides a crate up a ramp at angle of 27 deg. by exerting a 285 N force parallet to the ramp. The crate moves at a constant spped. The coefficient of friction is 0.23. How much work did this guy do when the crate was raised a vertical distance of 1.27 m?

I started with constant speep. It means that the force 285 N = mg*sin27 + mgcos27*0.23

m = 44.13 kg.

delta U = 44.13 * 9.8 * 1.27 = 549 J

W = - delta U = - 549 J.

The answer was wrong. Can anyone help. Thanks in advance.

 

[Xamfy19]

Thanks, I got

I just figured out that I should have used W = F * D since it is moving at constant speed. The combination of friction and gravity force will be equal to the F. That's all he need to move to move crate up.

The answer is 797.26 J

 

[M98Ranger]

Hi there, it seems like you have the right approach to solving this problem. However, there may be a mistake in your calculation of the weight of the crate. The weight should be equal to the mass multiplied by the gravitational acceleration (9.8 m/s^2), not just the mass itself. So the correct weight should be 432.774 N (44.13 kg * 9.8 m/s^2).

Also, when calculating the work done by the guy, you should use the component of the force parallel to the ramp, which is 285 * sin27 = 126.3 N. Using this force in your equation, the work done should be 160.701 J (126.3 N * 1.27 m).

Remember to always double check your calculations and units to avoid any mistakes. I hope this helps and good luck with your problem!