A spring of stiffness 2 kN/m is attached to a block of mass 8 kg on a surface that has frictional co-efficient 0.4. The spring is compressed and released, and the block accelerates at 4 m/s^2 just as it is released. The work done in compressing the spring is:
Ops:
A. 584 mj В. 1288 mj C. 1024 mJ D 672 mj
Answers
Answered by
2
the answer is :1024 and I know the answer
Answered by
2
Explanation:
Given A spring of stiffness 2 k N/m is attached to a block of mass 8 kg on a surface that has frictional co-efficient 0.4. The spring is compressed and released, and the block accelerates at 4 m/s^2 just as it is released. The work done in compressing the spring is
- There is a spring and the block. In the same position the block is moved a little. Let d be the distance moved by the block and also the length of the spring compressed.
- Now k = 2 KN / m = 2000 N/m
- Now when the block is just released the normal force acting will be mg.
- So N = mg
- = 8 x 10
- = 80 N
- So the friction will act backwards and k is pushing it away. The entire block is moving with an acceleration of 4 m/s^2. So we have spring force Fk and frictional force .
- So friction will be f = 0.4 x 80
- = 32 N
- Now Fk = k x d
- = 2000 d
- According to Newton’s formula
- Fnet = ma
- Fk – f = ma (overall block is mowing towards right)
- 2000 d -32 = 8 x 4
- 2000 d – 32 = 32
- 2000 d = 64
- d = 32 mm
- Work done in compressing the spring will be potential energy stored in spring
- W = ½ k d^2
- = 2000 / 2 x 32 x 32 / 10^6
- = 1024 / 1000
- W = 1024 mJ
Reference link will be
https://brainly.in/question/40709716
Similar questions