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A 75.5 kg crate moving on flat ground slows down and comes to a complete stop due to a frictional force.

The coefficient of kinetic friction is 0.262. What is the acceleration of the crate?

[tex]a = [?] \, \text{m/s}^2[/tex]

Sagot :

Sure, let's solve the problem step-by-step.

1. Identify the given values:
- Mass of the crate ([tex]\( m \)[/tex]): 75.5 kg
- Coefficient of kinetic friction ([tex]\( \mu_k \)[/tex]): 0.262
- Acceleration due to gravity ([tex]\( g \)[/tex]): 9.81 m/s²

2. Calculate the normal force ([tex]\( N \)[/tex]):
The normal force is the force exerted by a surface to support the weight of an object resting on it. On flat ground, this equals the weight of the object.
[tex]\[ N = m \times g \][/tex]
Substituting the given values:
[tex]\[ N = 75.5\, \text{kg} \times 9.81\, \text{m/s}^2 = 740.655\, \text{N} \][/tex]

3. Calculate the frictional force ([tex]\( F_{\text{friction}} \)[/tex]):
The frictional force can be calculated using the coefficient of kinetic friction and the normal force.
[tex]\[ F_{\text{friction}} = \mu_k \times N \][/tex]
Substituting the known values:
[tex]\[ F_{\text{friction}} = 0.262 \times 740.655\, \text{N} = 194.05161\, \text{N} \][/tex]

4. Determine the acceleration ([tex]\( a \)[/tex]):
According to Newton's second law, force is equal to mass times acceleration ([tex]\( F = m \times a \)[/tex]). The frictional force is what causes the crate to decelerate.
[tex]\[ F_{\text{friction}} = m \times a \][/tex]
Rearrange to solve for acceleration ([tex]\( a \)[/tex]):
[tex]\[ a = \frac{F_{\text{friction}}}{m} \][/tex]
Substituting the values:
[tex]\[ a = \frac{194.05161\, \text{N}}{75.5\, \text{kg}} = 2.57022\, \text{m/s}^2 \][/tex]

Since the frictional force is acting in the direction opposite to the motion of the crate, the acceleration is negative (indicating deceleration):
[tex]\[ a = -2.57022\, \text{m/s}^2 \][/tex]

Therefore, the acceleration of the crate is [tex]\( -2.57022 \, \text{m/s}^2 \)[/tex].