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kow [346]
3 years ago
13

A 2.0 kg block has a rope attached to the block on a table and is pulled with a force of 8.0 N. The block accelerated at 2.5m/s^

2. The coefficient of kinetic friction between the block and the table is ______. (Round to the nearest hundred.)
Physics
1 answer:
Masja [62]3 years ago
3 0

Answer:

0.15

Explanation:

Assuming the rope is horizontal, sum the forces in the y direction:

∑F = ma

N − mg = 0

N = mg

Sum the forces in the x direction:

∑F = ma

F − Nμ = ma

Substitute:

F − mgμ = ma

mgμ = F − ma

μ = (F − ma) / (mg)

Plug in values:

μ = (8.0 N − 2.0 kg × 2.5 m/s²) / (2.0 kg × 9.8 m/s²)

μ = 0.15

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Observe: Click Reset. Turn on Show velocity vector and Show velocity components. Set vinitial to 50 m/s and set θ to 45.0 degree
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v_y = v_{oy} - g t

where the upward direction is positive, so the arrow represents this speed (blue) must decrease, reach zero and grow in a negative direction as time progresses

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In this exercise you are asked to observe the change in velocity in a projectile launch.

If we assume that the friction force is small, the velocity in the x-axis must be constant

         vₓ = v₀ₓ

Therefore, the arrow (red) that represents this movement must not change in magnitude.

In the direction of the y axis, the acceleration of gravity is acting, so the magnitude of the velocity in this axis changes

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where the upward direction is positive, so the arrow represents this speed (blue) must decrease, reach zero and grow in a negative direction as time progresses

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