Anonymous
Anonymous asked in Science & MathematicsPhysics · 3 weeks ago

# 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?

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• MyRank
Lv 6
2 weeks ago

Given,

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.

• Mangal
Lv 4
3 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)

• oubaas
Lv 7
3 weeks ago

m/2*V^2 = F*d

d = 2*8^2/35 = 128/35 = 3.66 m

• 3 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.

• 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

= 3.7m

• 3 weeks ago

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