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14 days ago

15

A 250-kg crate is on a rough ramp, inclined at 30° above the horizontal. The coefficient of kinetic friction between the crate a

nd ramp is 0.22. A horizontal force of 5000 N is applied to the crate, pushing it up the ramp. What is the acceleration of the crate?

2 answers:

inna [987]14 days ago

6 0

Answer:

$8.35 m/s^2$

Solution Explanation:

Given data:

Mass of the crate = 250 kg

$\theta=30^{\circ}$

Kinetic friction coefficient between crate and ramp = 0.22

Applied horizontal force = 5000 N

Objective: Calculate the crate's acceleration.

The normal force acting perpendicular to the ramp surface is:

$=N=Fsin\theta+mgcos\theta=5000sin 30^{\circ}+250\times 9.8\times cos30^{\circ}=4623.93 N$

The friction force opposing the crate's motion is calculated as:

$0.22\times 4623.93=1017.26 N$

According to Newton's second law, the net force causing the acceleration is:

$ma=Fcos\theta-(F_f+mgsin\theta)$

$a=\frac{5000cos 30^{\circ}-(1017.26+250\times 9.8 sin30^{\circ}}{250}$

$a=8.35 m/s^2$

Therefore, the crate’s acceleration is: $8.35 m/s^2$

serg [1.1K]14 days ago

6 0

The force moving the crate upward along the incline acts against at least two other forces: gravitational force pulling it downwards and friction between the crate and the ramp’s rough surface.

Force on any object can be described through its mass and acceleration.

The kinetic friction force is determined by the relation between applied force and friction force, taking gravity into account.

So, 5000 minus 22% of 5000 equals 250 multiplied by acceleration.

Therefore, acceleration equals 3900 divided by 250 meters per second squared.

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