a 4 kg box is travelling at a speed of 8 m/s.?
a 4 kg box is travelling at a speed of 8 m/s. Suddenly it comes across a rough patch which exerts a friction force of 35 N. How far will the box go before it stops?
- MyRankLv 62 weeks agoFavorite Answer
Mass (m) = 4kgSpeed (v) = 8m/secForce (F) = 35NDistance (d) = ?We know that:-Work done (w) = force (F) x distance (d)= ½mv² = 35N x distance (d)= mv² = 2 x 35 x distance (d)= 4 x (8)² = 70 x distance (d)= distance (d) = 256/70 = 3.65m∴ Distance (d) = 3.65m.Source(s): http://myrank.co.in/
- MangalLv 43 weeks ago
Kinetic energy of the box = (1/2) (4) (8)² = 129 Joule
The frictional force will have to do above work in order to stop the box (making KE = 0)
work = force x distance
distance = work / force = 128 / 35 = 3.657 m (Ans)
- oubaasLv 73 weeks ago
m/2*V^2 = F*d
d = 2*8^2/35 = 128/35 = 3.66 m
- MorningfoxLv 73 weeks ago
Force = mass * acceleration = 35 N = 4 kg * a
a = 35 / 4 (m/s^2)
time to stop = (8 m/s) / a = 0.914286 seconds.
distance = 0.5 * a * t^2 = 3.657 meters. Round off as appropriate.
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- 3 weeks ago
OK so the Work that must be done to bring it to rest = F * d.
Work = kinetic energy of the box
Ek = 1/2 m * v^2 = F * d
d = (1/2 * m * v^2) / 35N
= (1/2 * 4kg * (8 m/s)^2) / 35N
- 3 weeks ago