Question

A block of mass 102kg is moving with a velocity 53m/s on horizontal. Block is stopped by applying a deceleration of 2.0m/s^2 find:
(a) Applied force to stop block
(b) Work done by deceleration force to stop block
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Answer 1

We have,

Mass of block = 102 kg

Initial velocity = 53 m/s

Final velocity = 0 m/s (rest)

Decceleration = 2 m/s²

(a) We know, F = m × a

⇒ F = (102 kg)(2 m/s²)

⇒ F = 204 kg m/s² = 204 N (1).

(b) We also know, 2as = v² - u²

⇒ - 2as = - u² (-ve acceleration)

⇒ s = u²/2a = (53 m/s)²/(2 × 2 m/s²) = 53 × 53/4 m

⇒ s = 702.25 m

∵ W = F s cosθ

⇒ W = 204 N × 702.25 m × cos180°

⇒ W = 143259 Nm × - 1 (since cos180° = - 1)

⇒ W = - 143259 J = - 143.259 kJ (2).

Answer 2

$\bold{Solution\;:-}$

Given :-

› Mass of the block l, m = 102 kg

› Initial velocity by which it was moving, u = 53 m/s

› Final velocity [ when it stopped it came to rest ], v = 0 m/s

› -ve acceleration, a = 2 m/s²

a. By using the formula, F = m*a

F = 102 * 2 = 204 N [ in backward direction ]

b. Work done by deceleration force to stop block, W = Force*s*cos θ

› By using the equation of motion = 2as = v²- u²

› 2(-2)s = (0)² - (53)²

› - 4*s = - 2809

› s = 2809/4

› s = 702.25 m

Therefore, W = 204*702.25*cos θ

› W = 143259*cos 180°

› W = 143259* - 1

› W = - 143259 J

› W = - 143.25 kJ

Hope it helps!