Work Energy and Power Test - 11

As, the magnetic field due to motion of electron and proton act in a direction perpendicular to the direction of motion, no work is done by the forces. This is why one ignores the magnetic force of one particle on another.

Force between two protons = force between a proton and a positron. Because of having much lighter weight than proton, positron moves away a larger distance compared to proton. 

As, work done = force x displacement, therefore in the same time t, work done in case of positron is more than that of proton.

In the whole process, the man exerts a variable force (F) on the ground to set his body in motion. This force is in addition to the force required to support his weight (mg). Once the man is in standing position, F become zero.

Just because road does not move at all so the work done by the cycle on the road must be zero.

\(F={\Delta p\over t}={mv-(-mv)\over t}\)

= \(2mv\over t\)

= \(2\times 150\times 10^{-3}Kg\times35m/s\over 0.001s\)

= 10500 N

= \(1.05 \times 10^4\, N\)

All forces are internal force, gravitational force and frictional force.

Work done by gravitational force = –mgL

Work done by internal force

= −work done against gravitational force = mgL

Work done by frictional force = F x 0 = 0

Work done by all forces = mgL – mgL + 0 = 0

Work done by reaction force = N x 0 = 0

Given

m = 0.5Kg, v = k\(x^{3/2}\) where k = \(5m^{–1/2}s^{–1}\)

(Here k is used as constant a)

Acceleration, \(a={dv\over dt}={dv\over dx} \cdot {dx\over dt}={vdv\over dx}\)       ...(i)

Differentiating (i) with respect to x,

\(2v{dv\over dx}=3k^2x^2\)     \([\because\,\,v^2=k^2x^3]\)

\(\therefore\) Acceleration, \(a={3\over2}k^2x^2\)

Force, F = Mass x Acceleration = \({3\over2}mk^2x^2\)

Work done, w = \(\int\) F dx = \(\int\limits_0^2 {3\over2}mk^2x^2dx\)

W = \({3\over2}mk^2[{x^3\over 3}]^2_0\)

= \({3\over6}\times 0.5\times 5^2\times [2^3-0]\)

= 50 J

Suppose, m = mass of the body

Given :

a = acceleration produced in the body

v = velocity of the body

P = Power delivered to the body in a time t

Using Newton’s second law of motion, the force is given by

F = ma

Both m and a are constants.

Hence, force will be a constant

F = ma = constant             ...(i)

For velocity v, acceleration is given by

\(a={dv\over dt}\)

v = at

p = Fv

P = \(a^2mt\)

or

Since, a and m are constants, therefore

P = constant x t

or \(P\propto t\)

A body is falling under the gravity then kinetic energy increases and potential energy decreases but total mechanical energy (Kinetic energy + Potential energy) is constant.

In inelastic collision total linear momentum remains conserved. But total energy does not remain conserved.

Given mass, m = 5 kg,

Radius, R = 1 m

v =300rpm = \(360\over60\)rps = 5 rps

The angular speed,

\(\omega=2\pi v=2\pi \times 5=10\pi \,rad/s\)

The linear speed is

v = \(\omega R=(10\pi)(rad/s)(1\,m)\)

= \(10\pi\,m/s\)

K.E. = k = \({1\over2}mv^2\)

k = \({1\over2}\times (5kg)(10\pi\, m/s)^2\)

= \(250\pi^2\,J\)

Initial K.E. = \({1\over2}mv^2\)

= \({1\over2}(10Kg)\times (1m/s)^2\)

= 5 J

Initial PE at height 1.5m

= mgh

= \((10Kg)(10m/s^2)(1.5m)\)

= 150 J

Total initial energy

= 155 J

(a) \(K={1\over2}mv^2\) or \(v=\sqrt{{2K\over m}}\)

or \(v\propto \sqrt K\), at constant mass

\(\therefore\) \({v_1\over v_2}=\sqrt{K_1\over K_2}=\sqrt{2K\over K}=\sqrt2\)

or \(v_2={v_1\over \sqrt2}=0.707v_1\)

\(\therefore\) (a) is incorrect

(b) \(\because\) \(v_2=0.707v_1> 0.5v_1\)

so (b) is correct

(c) velocity of the bullet changed after hitting the target so it follows different parabolic path.

So, (c) is incorrect but (d) is correct.

(e) Bullet will follow parabolic path as it has both horizontal as well as vertical velocity after emerging from target, so (e) is incorrect.

(f) is correct because loss in KE is used to increase the internal energy of particles of the target.

Suppose constant power acts on the body of the mass m for a time t to give it a velocity v

K.E. = work done = power x time

or \({1\over2}mv^2=Pt\)

or v = \(\sqrt{2Pt\over m}\)      ...(i)

We know that \(v={dx\over dt}\)

or  dx = vdt

If x be the displacement of the body, then

= \(\int\)dx = \(\int\)vdt = \(\sqrt{2P\over m}\) \(\int\)\(t^{1/2}\) dt

= \(\sqrt{2P\over m}\Bigg({t^{1/2+1}\over {1\over2}+1}\Bigg)\)

= \(\sqrt{2P\over m}{t^{3/2}\over 3/2}={2\over3}\sqrt{2P\over m}t^{3/2}\)

Here, P = constant and m is also constant for a body,

so x = constant \(\times\) \(t^{3/2}\) or x is directly proportional to \(t^{3/2}\).