(33-34) a horizontal spring is compressed and a 4 kg block is placed on the left of the spring and a 7 kg block on the right. after the system is released from rest the 4 kg block slides to the left at 14 m/s and the 7 kg block slides to the right. the surface is frictionless. 33. the speed of the 7 kg block after it loses contact with the spring is a) 4 m/s b) 6 m/s c) 8 m/s d) 10 m/s e) 12 m/s

Speed, v₂ = 8 m/s

Explanation:

It is given that,

Mass of first block, m₁ = 4 kg

Mass of second block m₂ = 7 kg

Initially, the system is at rest, u₁ = u₂ = 0

Velocity of 4 kg block, v₁ = -14 m/s

Let v₂ is the velocity of the 7 kg block after it loses contact with the spring. Using the conservation of momentum as :

So, the speed of the 7 kg block after it loses contact with the spring is 8 m/s. Hence, this is the required solution.

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards