Answer :
Answer:
(a) K.E = 56000 J = 56 KJ, (b) d = 116.618 m
Explanation:
Given:
m = 70 Kg, Vi = 40.0 m/s, Vf= 0 m/s, μk = 0.70
Solution:
(a) K.E. =? J (due to motion of the runner the mechanical energy loss is in the form K.E.)
K.E. = 1/2 m v² = 0.5 ×70 kg × (40.0 m/s)²
K.E = 56000 J = 56 KJ
(b) distance d =? m
W= F × d
∴W = K. E = 56000 J and F= mg μk
K.E. = mg μk × d
so 56000 J = 70 kg × 9.8 m/s² × 0.70 × d
d = 116.618 m
Given Information:
Initial speed = vi = 40 m/s
Final speed = vf = 0 m/s
Mass of runner = m = 70 kg
Coefficient of friction = k = 0.70
Required Information:
a) Energy lost = ?
b) Distance = ?
Answer:
a) Energy lost = 56000 Joules
b) Distance = 116.6 meters
Explanation:
As we know the kinetic energy is given by
KE = 0.5mvi²
KE = 0.5*70*(40)²
KE = 56000 J
The work done by the runner is given by
W = Fd
We also know that the work is done in the form of KE
KE = Fd
where F = mgk substitute in the above equation so equation becomes
KE = mgkd
d = KE/mgk
where g = 9.8 m/s² is acceleration due to gravity and and k = 0.70 is friction coefficient.
d = 56000/70*9.8*0.70
d = 116.6 m
Therefore, the runner slide for 116.5 m after he stopped at the base.