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
V^2=u^2+2as
V=0
a =-u^2/2s
a=[4]^2/2[4]
a=-2m/s^2
The ration of output work to input work expressed as a percentage is called <u>Efficiency</u>.
Answer : The final energy of the system if the initial energy was 2000 J is, 3500 J
Solution :
(1) The equation used is,

where,
= final internal energy
= initial internal energy
q = heat energy
w = work done
(2) The known variables are, q, w and 
initial internal energy =
= 2000 J
heat energy = q = 1000 J
work done = w = 500 J
(3) Now plug the numbers into the equation, we get

(4) By solving the terms, we get




(5) Therefore, the final energy of the system if the initial energy was 2000 J is, 3500 J