Q#1
The mass of cyclist together with the bike is 90 kg. Calculate the increase in kinetic energy if the speed increases from 6.0 km/h to 12 km/h.Answer:
Initial speed = 6.0 km/h = 6000 m/3600 s = 5/3 m/s
K.E. = ½mv² = ½(90)(5/3)² = 45(25/9) = 125 J
Final speed = 12 km/h = 10/3 m/s
K.E. = ½(90)(10/3)² = 45(100/9) = 500 J
So, increase in K.E. = 500 J – 125 J = 375 J
Q#2
A block of mass 2.00 kg moving at a speed of 10.0 m/s accelerates at 3.00 m/s² for 5.00 s. Compute its final kinetic energy,
Answer:
Let us first compute velocity at the end of 5 s acceleration.
Here Initial velocity = u = 10.0 m/s, acceleration a = 3.00 m/s ² and time t = 5.00 s.
Final velocity v = u + at = 10 + 3(5) = 25 m/s
So final K.E. = ½mv² = ½(2)(25)² = 625 J.
Q#3
A box is pushed through 4.0 m across a floor offering 100 N resistance. How much work is done by the resisting force?
Answer:
Resisting force F = 100 N
Work done by this resisting force = 100 N(4 m) = 400 J
Q#4
A block of mass 5.00 kg slides down an incline of inclination 30° and length 10 m. Find the work done by the force of gravity.
Answer:
Force of gravity = mg = 5(9.8 N) = 49 N (Downward)
Displacement along the direction of force of gravity
= 10m(sin30°) = 5 m
So work done by the force of gravity = 49 N(5 m) = 245 J
Q#5
A constant force of 2.5 N accelerates a stationary particle of mass 15 g through a displacement of 2.50 m. Find the work done and the average power delivered.
Answer:
Initial velocity u = 0, Initial K.E. = 0
Acceleration a = Force/mass = 2.5/0.015 m/s² = 2500/15 m/s²
= 500/3 m/s² (Mass m = 15/1000 = 0.015 Kg)
Distance s =2.50 m
Final velocity v will be calculated using v² = u² + 2as
v² = 0 + 2(500/3)(2.50) = 1000(2.50/3) = 2500/3
Final K.E. = ½mv² = ½(15/1000)(2500/3) = 6.25 J
Work done = Change in K.E. = 6.25 J – 0 = 6.25 J
Let us calculate the time t taken through this displacement.
From v = u + at
√(2500/3) = 0 + (500/3)t
50/√3 = 500t/3
t = 0.1√3 s
Average Power = Work Done/time
= 6.25/(0.1√3)
= 36.08 W
Q#6
A particle moves from a point r1 = (2m)i+ (3m)jto another point r2 = (3m)i + (2m)j during which a certain force F = (5N)i + (5N)j acts on it. Find the work done by the force on the particle during the displacement.
Answer:
Displacement of the particle
∆r = r2 – r1
= (3m)i + (2m)j – {(2m)i + (3m)j}
= (1 m)i – (1 m)j
Work Done = Dot Product of vectors F and ∆r.
W = F.∆r
= {(5N)i + (5N)j}{(1m)i – (1 m)j}
= –5 J + 5 J
= 0
Q#7
A man moves on a straight horizontal road with a block of mass 2 kg in his hand. If he covers a distance of 40 m with an acceleration of 0.5 m/s², find the work done by the man on the block during the motion.
Answer:
Velocity v after 40 m is given by, v² = u² + 2as
(Where acceleration a = 0.5 m/s², Distance covered s = 40 m)
v² = 0 + 2(0.5)40 = 40
Work done = Change in K.E.
= ½mv² – ½mu²
= ½(2)(40) – 0
= 40 J
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