183k views
20 votes
A dock worker pushes a 50 kg crate up a 1-m-high, 3-m-long ramp.

Ignoring friction, how much work did he do?

Pls help !!!!!

User Prd
by
4.5k points

1 Answer

5 votes

Final answer:

The work done by the dock worker pushing the crate up the ramp is 490 J.

Step-by-step explanation:

The work done by a dock worker in pushing a 50 kg crate up a 1-m-high ramp can be calculated using the formula:



Work = Force x Distance x Cos(theta)



Since the ramp is 3 m long and the height is 1 m, we can calculate the distance (d) using the Pythagorean theorem:



d = sqrt(3^2 + 1^2) = sqrt(10) = 3.16 m



Assuming the ramp is at an angle of θ with respect to the horizontal, the work can be calculated as:



Work = Force x Distance x Cos(theta) = mgh x Cos(theta)



where m is the mass of the crate (50 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the ramp (1 m).



Plugging in the values, we get:



Work = 50 kg x 9.8 m/s^2 x 1 m x Cos(theta)



Since we are ignoring friction, the angle of the ramp is irrelevant, and Cos(theta) is 1. Therefore, the work done is:



Work = 50 kg x 9.8 m/s^2 x 1 m x 1 = 490 J

User DAC
by
4.8k points