Question

A force of 500 N is acting on a block of 50 kg mass resting on a horizontal surface as shown in the figure. Determine the velocity after the block has travelled a distance of 10m. Co efficient of kinetic friction is 0.5.

Answer 1

Given : Co-efficient of kinetic friction (μk)=0.5P = 500 N
m = 50 kg
u = 0 m/s
s = 10 m
To find : Velocity after the block has travelled a distance of 10 m

The body has no motion in the vertical direction.
∴ ΣFy = 0
∴ N - 50g + Psin30 = 0
∴N = 50g - 500sin30
Let us assume that F is the kinetic frictional force
∴ F = μk x N
∴F = 0.5(50 g – 500 sin 30)
∴F = 25g – 125

By Newton’s second law of motion
Σ Fx = ma ∴Pcos ϴ – F = 50a
∴50a = 312.7627
∴ a= 6.2553 m/s2
By kinematics equation
v2 = u2 + 2 x a x s
∴v2 = 02 + 2 x 6.2553 x 10
∴ v= 11.1851 m/s
The velocity of the block after travelling a distance of 10 m = 11.1851 m/s