Q#18
A 250 g block slides on a rough horizontal table. Find the work done by the frictional force in bringing the block to rest if is initially moving at a speed of 40 cm/s. If the frictional coefficient between the table and the block is 0.1, how far does the block move before coming to rest?Answer:
Mass m = 250 g = 0.25 kg
Initial speed u = 40 cm/s = 0.40 m/s
Initial K.E. = ½mu² = 0.5 x 0.25 x (0.40)2 = 0.02 J
Final K.E. = 0
Change in K.E. = 0.02 J = Work done by the frictional force
Frictional force = µmg = 0.1 x 0.25 x 9.8 N = 0.245 N
Let the block moves distance d before coming to rest,
so, Fd = 0.02
d = 0.02/0.245 = 8.20 cm
Water falling from 50 m high fall is to be used for generating electric energy. If 1.8x105 kg of water falls per hour and half the gravitational potential energy can be converted into electric energy, how many 100 W lamps can be lit?
Answer:
Mass of water falling per second = 1.8 x 105 kg/3600 s = 50 kg/s
Potential energy of falling water per second = mgh = 50 kg x 9.8 m/s² x 50 m = 34500 J
So electric energy generated =½ x 24500 J/s = 12250 W
Number of 100 W lamp that can be lit = 12250/100 = 122.5
Q#20
A person is painting his house walls. He stands on a ladder with a bucket containing paint in one hand and a brush in other. Suddenly the bucket slips from his hand and falls down on the floor. If the bucket with the paint had a mass of 6.0 kg and was at a height 2.0 m at the time it slipped, how much gravitational potential energy is lost together with the paint?
Answer:
Mass of the paint with bucket = m = 6.0 kg
Height h = 2.0 m
So, potential energy of the paint with the bucket in the person's hand
= mgh = 6.0 kg x 9.8 m/s² x 2.0 m = 117.6 J ≈ 118 J
Q#21
A projectile is fired from the top of a 40 m high cliff with an initial speed of 50 m/s at an unknown angle. Find its speed when it hits the ground.
Answer:
Potential Energy of the projectile with respect to the ground when it is projected = mgh = 40mg J
Kinetic Energy when projected = ½mv² = ½m(50)² = 1250m J
Total energy at the time of projection = K.E + P.E. = (40mg + 1250m) J
Let the speed of the projectile when it hits the ground = V, then at this point its P.E. = 0 (Since h = 0)
K.E. = ½mV² J
So, total energy = K.E. + P.E. = ½mV² J
Total Energy will remain constant.
So, ½mV²= 40mg + 1250m
V² = 80g + 2500 = 3284
V = 57.31 m/s
The 200 m freestyle women's swimming gold medal at Seol Olympic 1988 went to Heike Friendrich of East Germany when she set a new Olympic record of 1 minute and 57.56 seconds. Assume that she covered most of the distance with a uniform speed and had to exert 460 W to maintain her speed. Calculate the average force of resistance offered by the water during the swim.
Answer:
Let the force applied by swimmer = F and force of resistance offered by the water = R
So resultant force in the direction of motion = F – R
Since the swimmer swims with a uniform speed, resultant force is zero, i.e. F – R = 0, →F = R
Since she exerted 460 W of power, so water resistance also resisted with 460 W of power.
If v = uniform speed of the swimmer = 200 m/(60 + 57.56) s = 1.70 m/s
Now Power =Force x speed
460 = R x 1.70
R= 460/1.70 = 270.60 N
The US athlete Florence Griffith Joyner won the 100 m sprint gold medal at Seol Olympic 1988 setting a new Olympic record of 10.54 s. Assume that she achieved her maximum speed in a very short time and then ran the race with that speed till she crossed the line. Take her mass to be 50 kg. (a) Calculate the kinetic energy of Griffith Joyner at her full speed.(b) Assume that the track, the wind etc. offered an average resistance of 1/10th of her weight, calculate the work done by the resistance during the run.(c) What power Griffith Joyner had to exert to maintain uniform speed?
Answer:
Uniform speed of the Athlete v= 100 m/10.54 s = 9.49 m/s
Mass m = 50 kg
(a) Kinetic Energy of the Athlete at full speed = ½mv² = 0.5 x 50 x (9.49)² = 2251 J
(b) Weight of the Athlete = mg = 50 x 9.8 N = 490 N
Force of resistance = 1/10 of 490 N = 49 N
Since she runs with a constant speed, the Net force on her in the direction of movement is zero, i.e. Force exerted by her and the force of resistance are equal in magnitude and opposite in direction.
So,work done by the resistance = Force x distance
= 49 N x (–100 m) = –4900 J (Negative sign is for the movement opposite to the Force)
(c) Since force applied by the Athlete and the resistance are equal in magnitude but opposite in direction, So force applied F = 49 N,
Constant speed of the Athlete = v = 9.49 m/s
So power exerted by her = Fv = 49 x 9.49 W = 465 W
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