Q#34
A small block of mass 200 g is kept at the top of a frictionless incline which is 10 m long and 3.2 m high. How much work was required (a) to lift the block from the ground and put it at the top, (b) to slide the block up the incline? What will be the speed of the block when it reaches the ground if (c) it falls off the incline and drops vertically on the ground (d) it slides down the incline? Take g = 10 m/s².Answer:
(a) Mass of the block = 200 g = 0.200 kg
Height of incline h = 3.2 m, Length of the incline l = 10 m
Required work done to lift the block from the ground and put it at the top = Work done against the gravity
= mgh = 0.200(10)(3.2) J = 6.40 J
(b) Since the incline is frictionless, only force on the block is that of gravity. Even if it is slid on the incline work is done against the gravity. The movement against the gravity, in this case, is also same i.e. 3.2 m.
So, in this case also work done = 6.40 J
(c) When the block falls vertically downward, its speed near the ground
v² = u² + 2gh = 0 + 2(10)(3.2) = 64
v = 8 m/s
(d) When it slides down the incline, let its speed near the ground be v. Since the only force on the block is that of gravity, the W.D. by the gravity = change in its Kinetic Energy.
mgh = ½mv²
gh = v²/2
v² = 2gh
v = √(2gh) = √(2 x 10 x 3.2) = 8 m/s
Q#35
In a children's park, there is a slide which has a total length of 10 m and a height of 8.0 m (Figure 8-E3). A vertical ladder is provided to reach the top. A boy weighing 200 N climbs up the ladder to the top of the slide and slides down to the ground. The average friction offered by the slide is three tenth of the weight. Find (a) the work done by the ladder on the boy as he goes up, (b) the work done by the slide on the boy as he comes down. (c) Find the work done by forces inside the body of the boy.
Answer:
(a) When the boy goes up the ladder applies a normal force on the boy which is equal to the weight of the boy = mg = 200 N. Since the movement and the force both are in the same direction the work done is positive. The height of the ladder = 8 m
Hence the work done = 200 N x 8 m = 1600 J.
(b) The slide does the work on the boy through friction. Frictional force = 3/10th of the weight = (3/10)200 N = 60 N
Length of the slide = 10 m
So, work done = 60 N x (–10)m = –600 J
(Negative sign is due to the movement against the force)
(c) Work done by the forces inside the body of the boy will be zero because inside the body the forces (internal force) will be in equal and opposite pairs. So, net internal force = 0.
Q#36
Figure (8-E4) shows a particle sliding on a frictionless track which terminates in a straight horizontal section. If the particle starts slipping from the point A, how far away from the track will the particle hit the ground?
Answer:
It is clear from the figure that at the point of termination of the track the particle descends through 1.0-0.5 = 0.5 m
Its initial K.E. is zero. At the point of termination of the track, its K.E. will be due to change in P.E. Let its velocity at this point be v.
K.E. = ½mv²
Change in P.E. = mgh, Equating them we get,
½mv² = mgh
v²/2 = gh
v = √(2gh) = √{2(9.8)0.5} = 3.13 m/s
Now the particle will travel in a parabolic path, The vertical distance traveled = s = 0.50 m, with initial velocity = 0. Let t be the time to reach the ground. Then from s = ut + ½gt² we get,
0.5 = 0 + 0.5(9.8)t²
t² = 1/9.8
t = 0.32 s
In this time the distance traveled = speed x time
= 3.13 x 0.32 = 1.00 m from the end of the track.
Q#37
A block weighing 10 N travels down a smooth curved track AB joined to a rough horizontal surface (figure 8-E5). The rough surface has a friction coefficient of 0.20 with the block. If the block starts slipping on the track from a point 1.0 m above the horizontal surface, how far will it move on the rough surface?
Potential energy of the block at Point A with respect to ground = mgh = 10 N x 1.0 m = 10 J
K.E. at A = 0
Total Mechanical Energy at A = P.E. + K.E, = 10 J
P.E. at B = 0
K.E. at B = ½mv² (v = speed at B and mass m = 10/g kg)
Total Mechanical Energy at B = ½mv² = ½(10/g)v² = 5v²/g
Total Mechanical energy will be constant.
So, ½(10/g)v² = 10
v² = 2g
Now on the rough surface, the force of friction will oppose the motion. Frictional force F = µN, Where N is the normal force on the block which is equal to the weight.
N = 10 Newton and given that µ = 0.20
So, F = 0.20 x 10 = 2 N
Let us assume that it moves a distance d on the rough track. Then work done by the friction force = Fd = 2(–d) J = –2d J
Since only friction force is acting on the block along the motion, change in K.E. of the block = work done by the friction.
Change in K.E. = 0 – ½mv² = –½mv²
Equating them we get,
–2d = –½mv²
d = mv²/4 = (10/g)(2g/4) = 5.0 m
A uniform chain of mass m and length l overhangs a table with its two-third part on the table. Find the work to be done by a person to put the hanging part back on the table.
Answer:
Unit mass of the chain = m/l
Mass of hanging portion = (l/3)(m/l) = m/3
C.G. of the hanging part below the table = ½(l/3) = l/6
So, work to be done by the person = (m/3)g(l/6) = mgl/18
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