Answer: 321 J
Explanation:
Given
Mass of the box 
Force applied is 
Displacement of the box is 
Velocity acquired by the box is 
acceleration associated with it is 

Work done by force is 

change in kinetic energy is 

According to work-energy theorem, work done by all the forces is equal to the change in the kinetic energy
![\Rightarrow W+W_f=\Delta K\quad [W_f=\text{Work done by friction}]\\\\\Rightarrow 375+W_f=54\\\Rightarrow W_f=-321\ J](https://tex.z-dn.net/?f=%5CRightarrow%20W%2BW_f%3D%5CDelta%20K%5Cquad%20%5BW_f%3D%5Ctext%7BWork%20done%20by%20friction%7D%5D%5C%5C%5C%5C%5CRightarrow%20375%2BW_f%3D54%5C%5C%5CRightarrow%20W_f%3D-321%5C%20J)
Therefore, the magnitude of work done by friction is 
1 Newton in Earth gravity is the equivalent weight of 1/9.80665 kg on Earth. This is derived using Newton's second law f=ma and assuming Earth gravity of 9.80665 m/s2. 1 N (Earth) = 0.101971621297793 kg.
Answer:
The necessary separation between the two parallel plates is 0.104 mm
Explanation:
Given;
length of each side of the square plate, L = 6.5 cm = 0.065 m
charge on each plate, Q = 12.5 nC
potential difference across the plates, V = 34.8 V
Potential difference across parallel plates is given as;

Where;
d is the separation or distance between the two parallel plates;

Therefore, the necessary separation between the two parallel plates is 0.104 mm
Answer:
40 N/m
Explanation:
The diagram attached is used to answer the question
We know from Hooke's law that extension is directly proportional to the applied force hence
F=kx where x is extension, F is applied force and k is the spring constant. Making k the subject of the formula then

From the attached diagram extension is given by subtracting unstretched spring from stretched spring hence extension, x=1-0.5=0.5m
Substituting 20 N for F and 0.5 m for x then
