A train is moving with a velocity of 20 m/s. when the brakes are applied the acceleration is reduced to -0.6 m/s².calculate the distance covered by the train before coming to rest

x = 333.33 [m]

Explanation:

To solve this problem we must use the following kinematics equation.

where:

Vf = final velocity = 0

Vi = initial velocity = 20 [m/s]

a = desacceleration = 0.6 [m/s^2]

x = distance [m]

Note: the final speed is zero as the body finishes its movement.

Now replacing:

0 = (20)^2 - (2*0.6*x)

1.2*x = 400

x = 333.33 [m]

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

i think the answer is a and b

i wish you luck