The velocity of a body of mass 4.56 kg increases from \((10 ms^{-1}, 060^o) to (50 ms ^{-1}, 060^o)\) in 16 seconds . Calculate the magnitude of force acting on it.
- 17.1 N
-
11.4 N
- 36.5 N
- 5.7 N
The correct answer is: B
Explanation
\(m=4.56kg;t=16s;v_1=(10ms^{-1},060^o);v_2=(50ms^{-1},060^o)\)
Notice that there is no change in the direction of the velocity
So,we don't need to write it in a vector form
\(a=\frac{∆v}{t}=\frac{50 - 10}{16}\)
\(a=\frac{40}{16}=2.5ms^{-2}\)
F = ma = 4.56 x 2.5
∴ F = 11.4 N