Question

A mass resting on a horizontal, frictionless surface is attached to one end of a spring; the other end is fixed to a wall. It takes 3.7 J of work to compress the spring by 0.20 m . If the spring is compressed, and the mass is released from rest, it experiences a maximum acceleration of 12 m/s2 . Find the value of

(a) the spring constant and
(b) the mass.

Answer 1

(a) the spring constant is 185 N/m

(b) the mass is 3.083 kg

Step-by-step explanation:

Given;

work done in compressing the spring, W = 3.7 J

extension of the spring, x = 0.20 m

acceleration of the spring, a = 12 m/s²

Part (a) the spring constant

W = ¹/₂kx²

where;

w is work done

k is the spring constant

x is the extension of the spring

k = 2W / x²

k = ( 2 x 3.7) / (0.2²)

k = 185 N/m

Part (b) the mass

From Newton's second law of motion;

F = ma

Also from Hook's law;

F = kx

Equate the two laws;

ma = kx

m = kx / a

m = (185 x 0.2) / (12)

m = 3.083 kg

Answer 2

(a) 185 N/m

(b) 3.083 kg

Step-by-step explanation:

(a)

Using,

E = 1/2ke²....................... Equation 1

Where E = work done to compress the spring, k = spring constant of the spring, e = compression of the spring.

make k the subject of the equation

k = 2E/e²............... Equation 2

Given: E = 3.7 J, e = 0.20 m

Substitute into equation 2

k = 2(3.7)/0.2²

k = 185 N/m.

(b)

Using,

F = ma.............. Equation 2

Where F = force applied to the spring, m = mass attached to the spring, a = acceleration of the spring.

But from hook's law,

F = ke................. Equation 3

substitute equation 3 into equation 2

ke = ma

make m the subject of the equation

m = ke/a................ Equation 4

Given: k = 185 N/m, e = 0.2 m, a = 12 m/s²

Substitute into equation 4

m = 185(0.2)/12

m = 3.083 kg