Question

50.0 N (fwd) is applied to a 10.0 kg crate as it slides forwards. The coefficient of friction is 0.400. a) Find the acceleration of the box.

Answer 1

Approximately
`1.08\; {\rm m\cdot s^(-2)}` (assuming that
`g = 9.81\; {\rm N\cdot kg^(-1)}` and that the surface is level.)

Step-by-step explanation:

Under the assumptions, forces on this object would include:

• Weight of the object (vertical, downwards,)
• Normal force from the ground (vertical, upwards,)
• Forward force (horizontal, forward,) and
• Friction (horizontal, backward.)

The acceleration of this object can be found in the following steps:

• Find the normal force between the ground and the object.
• Find the friction on the object given the normal force.
• Find the normal force on the object.
• Divide the normal force on this object by mass to find acceleration.

If the object is on level ground, forces in the vertical direction should be balanced. Thus, the normal force from the ground should be equal in magnitude to the weight of this object.

Let
`m = 10.0\; {\rm kg}` denote the mass of this object:

`\begin{aligned} (\text{normal force}) &= (\text{weight}) \\ &= m\, g \\ &= (10.0\; {\rm kg}) \, (9.81\; {\rm N\cdot kg^(-1)}) \\ &= 98.1\; {\rm N}\end{aligned}`.

Since this object is moving on the surface, magnitude of the friction on the object would be equal to the product of:

• the magnitude of the normal force between the two surfaces, and
• the coefficient of kinetic friction
  `\mu_{\text{k}}`.

Let forward be the positive direction in the horizontal component. Friction would be negative since it opposes the motion and points backward:

`(\text{friction}) = -(0.400)\, (98.1\; {\rm N}) = (-39.24)\; {\rm N}`.

Since forces in the vertical direction are balanced, the net force on this object would be equal to the vector sum of forces in the horizontal direction:

`\begin{aligned} & (\text{forward force}) + (\text{friction})\\ =\; & 50.0\; {\rm N} + (-39.24\; {\rm N}) \\ =\; & 10.76\; {\rm N} \end{aligned}`.

Divide the net force on the object by mass to find the acceleration of the objecct:

`\begin{aligned} (\text{acceleration}) &= \frac{(\text{net force})}{(\text{mass})} \\ &= \frac{10.76\; {\rm N}}{10.0\; {\rm kg}} \\ &\approx 1.08\; {\rm m\cdot s^(-2)}\end{aligned}`.

In other words, the acceleration of the object would be approximately
`1.08\; {\rm m\cdot s^(-2)}`.