Question

A block with mass m is placed against a spring (with spring constant k) on a frictionless plane inclined with angle θ with respect to the horizontal. The block is not attached to the spring, but the spring is initially compressed by Δx from equilibrium. The block is then released and moves a distance d >Δx up the incline before momentarily coming to the stop. Which statements about the change in the spring potential energy ΔUs and change in the mechanical energy ΔEmech of the system are correct?
a.ΔEmech = 0
b.ΔUs < 0
c.ΔEmech < 0
d.ΔUs = 0
e.ΔEmech> 0
f.ΔUs > 0

Answer 1

Answer:

b.$\Delta U_s$

e.$\Delta E_{mech}>0$

Explanation:

We are given that

Spring constant=k

Angle=$\theta$

Mass=m

Distance=d

Distance=$\Delta x$

Initial potential energy of spring=$\frac{1}{2}k(\Delta x)^2$

Final potential energy of spring=0

Final potential energy  of system=mgd

Initial mechanical energy of system=Initial K.E+Initial P.E=0+0=0

Final mechanical energy of the system=Kinetic energy+Potential energy=0+mgd

$\Delta E_{mech}=mgd-0=mgd$>0

$\Delta U_{s}=Final-initial=0-\frac{1}{2}k(\Delta x)^2=-\frac{1}{2}k(\Delta x)^2=$

Option b and e are true.