Question

An ideal massless spring can be compressed 1 m by a force of 100 N. This same spring is placed at the bottom of a frictionless inclined plane which makes an angle $\theta ={30}^{\circ }$ with the horizontal. A 10 kg mass is released from rest at the top of the incline and is brought to rest momentarily after compressing the spring 2 meters.
(a) Through what distance does the mass slide before coming to rest?
(b) What is the speed of the mass just before it reaches the spring?

• A. $2m,v=2\sqrt{5}m/s$
• B. $5m,v=2\sqrt{5}m/s$
• C. $5m,v=\sqrt{5}m/s$
• D. $10m,5m,v=3\sqrt{50}m/s$

Answer 1

The correct option is A $2m,v=2\sqrt{5}m/s$


Let total distance moved by the block is. $S=(l+2)m$ Where l is the distance moved by the block before touching the spring.
Now, work done by gravity on the block is $W_g=mgSsin\theta =10\times 10\times Ssin30J$
$W_g=50SJ$ (1)
Work done by spring on the block is $W_s=-\frac{1}{2}k{x}^2$ Here K=100 N/m and x=2m
$W_s=-\frac{1}{2}\times 100\times 4J=W_s=-200J$ (2)
Total work done $W=W_g+W_s$
W=(50 s - 200) J
Since change in K.E. of the block is zero as $W=\Delta K.E.$
$50s-200=0\Rightarrow S=4m$
(b) as S = l + 2
l = S - 2 = 2m
Work done by gravity over this path length is
$W_g=mg\times 2msin\theta =10\times 10\times 2\times \frac{1}{2}=100J\because W_g=\Delta K.E.$
$100=\frac{1}{2}m{v}^2-0$
$v^2=\frac{100\times 2}{10}=20$
$v=2\sqrt{5}m/s$