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Chapter 4: Problem 453

A body having a mass of \(0.5 \mathrm{~kg}\) slips along the wall of a semispherical smooth surface of radius \(20 \mathrm{~cm}\) shown in figure. What is the velocity of body at the bottom of the surface $?\left(\mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)$ (A) \(2 \mathrm{~m} / \mathrm{s}\) (B) \(2 \mathrm{~m} / \mathrm{s}\) (C) \(2 \sqrt{2} \mathrm{~m} / \mathrm{s}\) (D) \(4 \mathrm{~m} / \mathrm{s}\)

Short Answer

Expert verified
The velocity of the body at the bottom of the surface is \(2 m/s\).

Step by step solution

01

Calculate potential energy at the top of the surface

At the top of the semi-spherical surface, the body has no kinetic energy as it begins to slide (assuming it starts from rest). Therefore, the total mechanical energy at the top is equal to the gravitational potential energy (PE) at that point. Using the gravitational potential energy formula, we have: \[PE = mgh\] Where: PE = potential energy m = mass of the body = 0.5 kg g = acceleration due to gravity = 10 m/s² h = height = radius of the semi-spherical surface = 20 cm = 0.2 m (converted to meters for consistent units) Now, we can calculate the potential energy: \[PE = (0.5 kg)(10 m/s²)(0.2 m)\]
02

Determine kinetic energy at the bottom of the surface

At the bottom of the surface, the body reaches the lowest point possible, so its potential energy is zero at this point. According to the conservation of mechanical energy, the potential energy at the top should now be completely converted into kinetic energy (KE) at the bottom. The kinetic energy formula is: \[KE = \frac{1}{2} mv^2\] Where: KE = kinetic energy m = mass of the body = 0.5 kg v = velocity of the body at the bottom (this is what we need to find)
03

Equate potential and kinetic energies and solve for velocity

Now, we can equate the potential energy at the top with the kinetic energy at the bottom: \[mgh = \frac{1}{2} mv^2\] We can cancel the mass (m) from both sides and solve for the velocity (v): \[gh = \frac{1}{2} v^2\] Now, substituting the given values for g and h, we get: \[(10 m/s²)(0.2 m) = \frac{1}{2} v^2\]
04

Find the velocity at the bottom of the surface

To find the velocity, we need to solve for v from the equation in the previous step: \[2(10 m/s²)(0.2 m) = v^2\] Now, calculating the value of v: \[v = \sqrt{2(10 m/s²)(0.2 m)} = 2 m/s\] Therefore, the velocity of the body at the bottom of the surface is 2 m/s, which corresponds to option (A).

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Most popular questions from this chapter

Two bodies of masses \(m_{1}\) and \(m_{2}\) have equal kinetic energies. If \(P_{1}\) and \(P_{2}\) are their respective momentum, what is ratio of \(\mathrm{P}_{2}: \mathrm{P}_{1}\) ? (A) \(\mathrm{m}_{1}: \mathrm{m}_{2}\) (B) \(\sqrt{\mathrm{m}}_{2} / \sqrt{\mathrm{m}_{1}}\) (C) \(\sqrt{m_{1}}: \sqrt{m_{2}}\) (D) \(\mathrm{m}_{1}^{2}: \mathrm{m}_{2}^{2}\)

What is the velocity of the bob of a simple pendulum at its mean position, if it is able to rise to vertical height of \(18 \mathrm{~cm}\) (Take \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\) ) (A) \(0.4 \mathrm{~m} / \mathrm{s}\) (B) \(4 \mathrm{~m} / \mathrm{s}\) (C) \(1.8 \mathrm{~m} / \mathrm{s}\) (D) \(0.6 \mathrm{~m} / \mathrm{s}\)

A body is moved along a straight line by a machine delivering a constant power. The velocity gained by the body in time \(t\) is proportional to..... (A) \(t^{(3 / 4)}\) (B) \(t^{(3 / 2)}\) (C) \(t^{(1 / 4)}\) (D) \(t^{(1 / 2)}\)

Four identical balls are lined in a straight grove made on a horizontal frictionless surface as shown. Two similar balls each moving with a velocity v collide elastically with the row of 4 balls from left. What will happen (A) One ball from the right rolls out with a speed \(2 \mathrm{v}\) and the remaining balls will remain at rest. (B) Two balls from the right roll out speed \(\mathrm{v}\) each and the remaining balls will remain stationary. (C) All the four balls in the row will roll out with speed \(\mathrm{v}(\mathrm{v} / 4)\) each and the two colliding balls will come to rest. (D) The colliding balls will come to rest and no ball rolls out from right.

A force \(F=A y^{2}+B y+C\) acts on a body in the \(y\) -direction. The work done by this force during a displacement from \(\mathrm{y}=-\mathrm{a}\) to \(\mathrm{y}=\mathrm{a}\) is (A) \(\left[\left\\{2 \mathrm{Aa}^{3}\right\\} / 3\right]\) (B) \(\left[\left\\{2 \mathrm{Aa}^{3}\right\\} / 3\right]+2 \mathrm{ca}\) (C) $\left[\left\\{2 \mathrm{Aa}^{3}\right\\} / 3\right]+\left[\left\\{\mathrm{Ba}^{2}\right\\} / 2\right]+\mathrm{ca}$ (D) None of these.
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