Question

QUESTION 20

Answer 1

a) $0.4 m/s^2$

We can find the acceleration of the box by using the suvat equation:

$s=ut+\frac{1}{2}at^2$

where

s is the distance travelled

u is the initial velocity

t is the time

a is the acceleration

For the box in the problem,

s = 16 m

t = 9 s

u = 0 (it starts from rest)

Solving for a, we find the acceleration:

$a=\frac{2s}{t^2}=\frac{2(16)}{9^2}=0.4 m/s^2$

C) Tension force in the chain: 446 N

In order to find the normal force, we have to write the equation of the forces along the vertical direction.

We have 3 forces acting along this direction on the box:

- The normal force, N, upward

- The force of gravity, mg, downward (where m= mass of the box and g = acceleration of gravity)

- The vertical component of the tension of in the chain, $T sin \theta$, upward

So the equation of the forces along the vertical direction is

$N+Tsin \theta - mg = 0$ (1)

Along the horizontal direction, instead, we have the following equation:

$T cos \theta - \mu N = ma$ (2)

where

$T cos \theta$ is the horizontal component of the tension in the chain

$\mu N$ is the frictional force

a is the acceleration

From (1) we write

$N=mg-T sin \theta$

And substituting into (2),

$T cos \theta - \mu (mg-T sin \theta) = ma\\T cos \theta - \mu mg + \mu T sin \theta = ma\\T = \frac{ma+\mu mg}{cos \theta + \mu sin \theta}$

And substituting:

m = 100 kg

$\theta=25^{\circ}$

$\mu=0.46$

$a=0.4 m/s^2$

$g=9.8 m/s^2$

We find the tension in the chain:

$T = \frac{(100)(0.4)+(0.46)(100)(9.8)}{cos 25 + 0.46 sin 25}=446 N$

B) Normal force: 792 N

We can now find the normal force by using again equation (1):

$N+Tsin \theta - mg = 0$

And substituting:

T = 446 N

m = 100 kg

$\theta=25^{\circ}$

$g=9.8 m/s^2$

We find:

$N=mg-T sin \theta=(100)(9.8)-(446)(sin 25)=792 N$