Question

A 50kg crate is being pushed on a horizontal floor at a constant velocity. The coefficient of kinetic friction between the crate and the floor is 0.1. How much force is applied to the crate?

Answer 1

ч= 0.1
m= 50kg

$F_{f} =$чmg
= (0.1)(50)
= 5 N

Answer 2

To solve the problem we will draw the free body diagram first and then calculate the normal force.

The force applies to the crate is 49.05 Newton.

Given to us

• Mass of the crate, m = 50 kg
• coefficient of kinetic friction between the crate and the floor,          μ = 0.1.

Free Body Diagram

Let's draw the Free Body Diagram of the crate first, as shown below, in the image.

Now,

Sum of all vertical force = 0,

$\sum F_y = 0$

$\text{Weight of the crate} = \text{Normal Force}$

$m\times g = N\\N = m\times g\\N = 50 \times 9.81$

Sum of all the Horizontal force = 0,

$\sum F_x =0$

$\rm{Force = Frictional\ Force$

$F = \mu \times N\\F = 0.1 \times 50 \times 9.81\\F = 49.05$

Hence, the force applies to the crate is 49.05 Newton.

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