Question

a 70 kg man standing on ice throws a 3 kg body horizontally at 8 m/s. the friction coefficient between the ice and his feet is 0.02. the distance, the man slips is : (a) 0.3 m (b) 2 m (c) 1 m (d) ~

Answer 1

The distance at which the man slips is 0.3 m

Newton's Second Law, F = ma, is used to calculate the braking distance. By dividing the mass of the car by the gravitational acceleration, one may determine its weight. The weight of the car multiplied by the coefficient of friction equals the brake force.

Given-

mass of man= 70 kg

frictional coefficient μ=0.02

mass of body thrown= m2 = 3kg

let s be the stopping distance

we know that frictional force = F= μN

=μMg= 0.02 x 70 x 10

=14 N

∴acceleration, a= 14/70 = 0.2 m/s²

now on applying conservation of linear momentum

pi=pf            pi=0 (initially at rest)

0=m1v1-m2v2 (v1= velocity of man) (v2=velocity of body= 8m/s

v1= m2v2 /m1= 0.3 m/s

we know,

v²- u² = -2as

0- (0.3) ²= -2 x 0.2 x 5

s= 0.09/0.4 ≈ 0.3 m

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