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Answer 1

From the free body diagram, T−mg=5a

⇒T=5(9.8+2)=59N

The force T and the displacement are in the same direction.

Therefore, the Work done by tension =Tscos0∘=59×2.5×1=147.5J (here s = 2.5 m, the distance moved by the block)

The gravitational force mg and the displacement are in the opposite direction.

Work done against gravity =−mgs=5×9.8×2.5=−122.5J

Answer 2

33. (a) Work Done = F s cos Ф

Work Done = ( 5 × 9.8 ) N × 2 m × cos 180° [Since F = mg where m is  kg and g is 9.8 m per sq. s.

Work Done = 49 N × 2 m × ( -1 ) [ Since cos 180° = - 1 ]

Work Done = - 98 J

Since work done against gravity is considered positive therefore Work Done is 98 J.

(b) Kinetic energy = $\bold{ \frac{1}{2} mv^2}$

K.E = $\bold{\frac{1}{2} \times 1 \: kg \: v^2 = \frac{1}{2}v^2}$

Now, v = u + at

v = 0 + 10(5)

v = 50 m/s

K.E = 1/2 × 50 × 50 = 25 × 50 = 1250 J

(c) Power = Work Done / Time

1 kW = ( F s )/ t

1000 W = ( ma × s )/ t

1000 W = ( 500 × 10 × 40 )/t

1000 t = 200000

t = 200000 / 1000 = 200 seconds