Answer:
W = 7.06 J
Explanation:
From the given information the spring constant 'k' can be calculated using the Hooke's Law.

Now, using this spring constant the additional work required by F to stretch the spring can be found.
The work energy theorem tells us that the work done on the spring is equal to the change in the energy. Therefore,
![W = U_2 - U_1\\W = \frac{1}{2}kx_2^2 - \frac{1}{2}kx_1^2 = \frac{1}{2}(275.13)[0.29^2 - 0.18^2] = 7.06~J](https://tex.z-dn.net/?f=W%20%3D%20U_2%20-%20U_1%5C%5CW%20%3D%20%5Cfrac%7B1%7D%7B2%7Dkx_2%5E2%20-%20%5Cfrac%7B1%7D%7B2%7Dkx_1%5E2%20%3D%20%5Cfrac%7B1%7D%7B2%7D%28275.13%29%5B0.29%5E2%20-%200.18%5E2%5D%20%3D%207.06~J)
Answer:
a) pressure means force acted on a body per unit area
Power = work / time = 8000J / 20s = 400W
Answer:
a = 120 m/s²
Explanation:
We apply Newton's second law in the x direction:
∑Fₓ = m*a Formula (1)
Known data
Where:
∑Fₓ: Algebraic sum of forces in the x direction
F: Force in Newtons (N)
m: mass (kg)
a: acceleration of the block (m/s²)
F = 1200N
m = 10 kg
Problem development
We replace the known data in formula (1)
1200 = 10*a
a = 1200/10
a = 120 m/s²