Solutions to Exercises on Work and Energy HC Verma's Concepts of Physics Part 1 (44-48)

 Q#44

A block of mass m moving at a speed v Compresses a spring through a distance x before its speed is halved. Find the spring constant of the spring.  

Answer:
Initial Kinetic Energy of the bock = ½mv².
This energy will be conserved even after the spring compression.
Total energy at the point after spring compression of x =K.E. of the block + P.E. stored in the spring
= ½m(v/2)²  + ½kx². (where k is the spring constant)
Equating initial and final energies we get,

½m(v/2)² + ½kx² = ½mv²
mv²/4 + kx² = mv²
kx² = 3mv²/4
k = 3mv²/4x²

Q#45
Consider the situation shown in figure (8-E8). Initially, the spring is unstretched when the system is released from rest. Assuming no friction in the pulley, Find the maximum elongation of the spring.

Answer:
Let the maximum elongation of the spring be x. The block will also descend to x distance. So the reduction in P.E. of the block will be the P.E. stored in the spring.

Change in P.E. of the block = mgx = mgx
P.E. stored in the spring = ½kx²

Equating these two,

½kx² = mgx
X = 2mg/k

Q#46
A block of mass m is attached to two unstretched springs of spring constant k₁ and k₂ as shown in figure (8-E9). The block is displaced towards right through a distance of x and released. Find the speed of the block as it passes through the mean position shown.

Answer:
Let the speed of the block be v at the mean position. At the compressed position total energy is the P.E. stored in both the springs. i.e. equal to ½k₁x² + ½k₂x²

At the mean position, this P.E. will be changed to K.E. of the block. i.e. equal to ½mv².
Equating both we get

½mv² = ½k₁x² + ½k₂x²
mv² = k₁x² + k₂x²
v² = (k₁ + k₂)x²/m
v = √{(k₁ + k₂)/m}x

Q#47
A block of mass m sliding on a smooth horizontal surface with a velocity v meets a long horizontal spring fixed at one end and having spring constant k as shown in the figure (8-E10). Find the maximum compression of the spring. Will the velocity of the block be the same as v when it comes back to the original position shown?  

Answer:
Initial K.E. of the block = ½mv²
At the maximum Compression x, this K.E. will be changed to P.E. in the spring. This P.E. = ½kx²
Equating we get,

½kx² = ½mv²
kx² = mv²
x²= mv²/k
x = v√(m/k)

When it comes back to the original position its K.E. will again be the same (Assuming that there is no energy loss in the spring). Though the magnitude of the velocity of the block will be the same its direction will be opposite. Since velocity is a vector hence it is not the same as the original.

Q#48
A small block of mass 100 g is pressed against a horizontal spring fixed at one end to compress the spring through 5.0 cm (Figure 8-E11). The spring constant is 100 N/m. When released, the block moves horizontally till it leaves the spring. Where will it hit the ground 2 m below the spring? 

Answer:
Mass of the block m = 100 g = 0.10 kg, Spring constant k = 100 N/m, Compression of the spring x = 5 cm = 0.05 m

P.E. stored in the spring

= ½kx² = ½(100)(0.05)² = 0.125 J

This P.E. is converted as K.E. of the block when it leaves the spring = ½mv²
i.e.
½mv² = 0.125
v² = 0.25/m = 0.25/0.10 = 2.5
v = √2.5 m/s

Let time taken to fall 2 m be t. In vertical direction u = 0, h = 2 m

from h = ut + ½gt²
2 = 0 + ½(9.8)t² = 4.9t²
t² = 2/4.9
t = √(2/4.9) s

In this time the block will move horizontally with a uniform velocity of v = √2.5 m/s
So the horizontal distance traveled

= vt = (√2.5)(√2/4.9) m
= √5/4.9 ≈ 1 m

So the block will hit the ground at a distance of 1 m from the free end of the spring. 

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