The kinetic energy acquired by a mass m in travelling distance d, starting from rest, under the action of a constant force is directly proportional to

Question

The kinetic energy acquired by a mass m in travelling distance d, starting from rest, under the action of a constant force is directly proportional to (A) m (B) m0 (C) √ m (D) 1√ m

Tardigrade Admin · Accepted Answer

v2=u2+2 a s or v2-u2=2 a s or v2-(0)2=2 × (F/m) × s or v2=(2 F s/m) and K.E. =(1/2) m v2=(1/2) m × (2 F s/m)=F S. Thus K.E. is independent of m or directly proportional to m0.

Q. The kinetic energy acquired by a mass m in travelling distance $d$ , starting from rest, under the action of a constant force is directly proportional to

$m$

$m^{0}$

$m$

$1 m$

$v^{2} = u^{2} + 2 a s$
or $v^{2} - u^{2} = 2 a s$
or $v^{2} - (0)^{2} = 2 \times \frac{F}{m} \times s$
or $v^{2} = \frac{2 F s}{m}$
and K.E. $= \frac{1}{2} m v^{2} = \frac{1}{2} m \times \frac{2 F s}{m} = FS$ .
Thus K.E. is independent of $m$ or directly proportional to $m^{0}$ .

Q. The kinetic energy acquired by a mass m in travelling distance d, starting from rest, under the action of a constant force is directly proportional to

m

m0

m​

1m​

v2=u2+2as or v2−u2=2as or v2−(0)2=2×mF​×s or v2=m2Fs​ and K.E. =21​mv2=21​m×m2Fs​=FS. Thus K.E. is independent of m or directly proportional to m0.

Q. The kinetic energy acquired by a mass m in travelling distance $d$ , starting from rest, under the action of a constant force is directly proportional to

$m$

$m^{0}$

$m$

$1 m$

$v^{2} = u^{2} + 2 a s$
or $v^{2} - u^{2} = 2 a s$
or $v^{2} - (0)^{2} = 2 \times \frac{F}{m} \times s$
or $v^{2} = \frac{2 F s}{m}$
and K.E. $= \frac{1}{2} m v^{2} = \frac{1}{2} m \times \frac{2 F s}{m} = FS$ .
Thus K.E. is independent of $m$ or directly proportional to $m^{0}$ .