Answer to Question #171425 in Physics for Juan dela cruz

Question

A 2.70 kg block starts from rest and slides a distance d

down a frictionless 32.1° inclined surface (see figure).

While sliding, it comes into contact with an uncompressed 

spring of negligible mass and force constant of 760 N/m. If 

the height through which the block slides down the inclined 

is 2.40 m, (a) find the maximum compression of the spring.

(b) What is the speed of the mass when it comes into 

contact with the spring? (c) What is the initial separation 

distance d between the mass and the spring?

Answer 1

a)

$$0.5kx^2=mgh\\0.5(760)x^2=(2.7)(9.8)(2.4)\\x=0.409\ m$$

b)

$$0.5mv^2=mgh-\frac{mgx}{\sin{32.1}}\\ 0.5v^2=(9.8)(2.4)-\frac{(9.8)(0.409)}{\sin{32.1}}\\v=5.65\frac{m}{s}$$

c)

$$d=\frac{h}{\sin{32.1}}-x\\d=\frac{2.4}{\sin{32.1}}-0.409=4.11\ m$$