M1 exam question help

Question 7

A sledge has mass 30kg. The sledge is pulled in a straight line along horizontal ground by means of a rope. The rope makes an angle 20 degrees with the horizontal. The coefficient of friction between the sledge and the ground is 0.2. The sledge is modelled as a particle and the rope as a light inextensible string. The tension in the string is 150N. Find to 3sf;

a) the normal reaction of the ground on the sledge.
b) the acceleration of the sledge.

When the sledge is moving at 12ms^-1, the rope is released from the sledge.

c) find to 3sf, the distance travelled by the sledge from the moment when the rope is released to the moment when the sledge comes to rest.

Reply 1

Newton

a) Resolving vertically:

N+150sin20-30g=0

Therefore:

N=30g-150sin20=243N

b) Resolving horizontally:

Tcos20-F=ma

150cos20-µR=30a

Therefore:

a=((150cos20-(0.2)(243))/30)=3.08m(s^(-2))

c) First of all we need to find the retardation that the friction will produce on the book once it is released from the string:

F=ma => a=(F/m)=(μN/m). Since the force is opposing any further motion we, and a being retardation we treat it as a negative. Therefore:

a=(-0.2(243)/30)=-1.62m(s^(-2)

By using the following formulae and taking into consideration then v=0:

(v^2)=(u^2)+2as

=> (-u^2)/2a=s

Therefore:

s=(-(12^2))/(-1.62)=88.9m

Newton.

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