Evaluate velocity of the block:
A block of 2500N rest on horizontal surface. The coefficient of friction in between block and plane is 0.3. The block can be pulled by the force of 1000N acting at 30º angle to the horizontal. Find out the velocity of the block after it moves over the distance of 30m, starting from the rest.
Here ∑V = 0 but ∑H≠ 0, As ∑H is converted into ma
∑V = 0
R + 1000sin30° - W = 0, W = 2500N
R = 2000N ...(i)
∑H≠ 0
∑H = µR - 1000cos30° = 266.02N ...(ii)
As ∑H≠ 0
By the newtons third law of motion
F = ma
266.02 = (2500/g) × (v^{2} - u^{2})/2.s => v^{2} = u^{2} + 2as
u = 0, v^{2} = {266.02 × 2 × s × g}/2500
v^{2} = {266.02 × 2 × 30 × 9.71}/2500
v = 7.91m/sec .......ANS