Solution :
It is given that a system is released from rest where no slack is attached to the cable and the spring is unstretched.
Mass of the cart = 30 kg
Applying the conservation of energy of the system,
Loss of the gravitational potential energy (PE) = the energy stored in the spring.
i.e., [tex]mg \Delta h = \frac{1}{2}kx^2[/tex]
[tex]mg (x\sin 25^\circ)=\frac{1}{2}kx^2[/tex]
[tex](30 \times 9.81) (x\sin 25^\circ)=\frac{1}{2}(125)x^2[/tex]
x = 1.98
Therefore the distance travelled by the 30 kg cart is 1.98 m.
