Question 20
a 100 kg box is being pulled by a chain along a horizontal surface. the chain is at a 25° angle. the coefficient of friction between the box and surface is 0.46
the box starts from rest and travels 16 meters in 9 seconds. what is the acceleration, tension force and normal force for the box?
a) acceleration
b) normal force
c) tension force

acceleration = 0.4m/s2

a)

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

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:

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, , upward

So the equation of the forces along the vertical direction is

(1)

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

(2)

where

is the horizontal component of the tension in the chain

is the frictional force

a is the acceleration

From (1) we write

And substituting into (2),

And substituting:

m = 100 kg

We find the tension in the chain:

B) Normal force: 792 N

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

And substituting:

T = 446 N

m = 100 kg

We find:

the answer is a. at the equator