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34 votes
34 votes
A 5.00 kilogram block is pulled 2.0 meters along a

horizontal, frictionless surface using a force of 100
newtons at an angle of 20.0° above horizontal as shown
in the diagram below. Determine the amount of work done in the block as it is moved

User Arclight
by
3.0k points

2 Answers

29 votes
29 votes

Final answer:

To calculate the work done, you need to use the component of the force that is in the direction of the displacement. The work done is 187.938 joules.

Step-by-step explanation:

The student has requested to determine the work done on a block by a force as it is moved along a frictionless surface. When a force is applied at an angle to the direction of displacement, only the component of the force that is in the direction of displacement does work. The work done (W) by a constant force (F) is calculated using the formula W = F * d * cos(θ), where d is the displacement and θ is the angle between the force and the direction of displacement. Since the surface is frictionless, we do not need to consider work done by friction.



For the given scenario with a 100 N force at a 20.0° angle pulling a block 2.0 meters:

W = F * d * cos(θ)

W = 100 N * 2.0 m * cos(20.0°)

W = 200 N*m * cos(20.0°)

W = 200 N*m * 0.93969 (cosine of 20 degrees)

W = 187.938 N*m or J (since 1 N*m = 1 J)



This calculation shows that the work done on the block is 187.938 joules (J).

User Myobis
by
2.7k points
15 votes
15 votes

Answer:

188 J

Step-by-step explanation:

W = F * d * cosα

Now F = 100 N, d = 2.0 m, α = 20.0°

so

W = 100 * 2.0 * cos(20) = 188 J

User Carl Smotricz
by
3.1k points