Question

A stockroom worker pushes a box with mass 11.2 kg on a horizontal surface with constant speed of 3.50m/s. The coefficient of kinetic friction between the box and the surface is 0.20.

What horizontal force must the worker apply to maintain the motion(Answer in N)?
If the force in the previous part is removed, how far does the box slide before coming to rest(Answer in m)?

Answer 1

Answer:

3.125 m

Explanation:

We are given that

Mass of box=m=11.2 kg

Speed of box=u=3.5m/s

Coefficient of kinetic friction=$\mu_k=0.2$

Final velocity,v=0

a.We have to find the horizontal force applied by worker to maintain the motion.

According to question

Horizontal force=F=$f=\mu_kmg$

$g=9.8m/s^2$

Substitute the values

Horizontal force=$F=0.2\times 11.2\times 9.8=21.95 N$

b.According to work-energy theorem

$W=\frac{1}{2}mv^2-\frac{1}{2}mu^2$

$-\mu mg s=\frac{1}{2}(11.2)(0)^2-\frac{1}[2}(11.20)(3.5)^2$

$-\mu mg s=-\frac{1}{2}(11.2)(3.5)^2$

$0.2\times (11.2)\times 9.8\times s=\frac{1}{2}(11.2)(3.5)^2$

$s=\frac{11.2\times (3.5)^2}{2\times 0.2\times 11.2\times 9.8}$

$s=3.125 m$

Hence, the box slide before coming to rest=3.125 m

Answer 2

(a) The horizontal force the worker must apply to maintain the motion is 21.95 N.

(b) If the force is removed, the distance traveled by the box is zero since it is not accelerating.

The given parameters;

• mass of the box, m = 11.2 kg
• constant, v = 3.5 m/s
• coefficient of friction, μ = 0.2

The horizontal force the worker must apply to maintain the motion is calculated as follows;

$F_f = \mu F_n\\\\F_f = 0.2 \times 11.2 \times 9.8\\\\F_f = 21.95 \ N$

The distance traveled by the box before coming to rest is calculated as;

at constant speed, acceleration , a = 0

$v^2 = u^2 + 2as\\\\0 = 3.5^2 + 2(0)s\\\\s = 0$

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