Question

A sleigh is being pulled horizontally by a train of horses at a constant speed of 6.38 m/s. The magnitude of the normal force exerted by the snow-covered ground on the sleigh is 7.50 ✕ 103 N.
(a) If the coefficient of kinetic friction between the sleigh and the ground is 0.26, what is the magnitude of the kinetic friction force experienced by the sleigh?
N

(b) If the only other horizontal force exerted on the sleigh is due to the horses pulling the sleigh, what must be the magnitude of this force?
N

Answer 1

Answer:

(a). The kinetic friction force is 1950 N.

(b). The magnitude of force will be equal of friction force

Explanation:

Given that,

Constant speed = 6.38 m/s

Force $F=7.50\times10^{3}\ N$

Kinetic friction = 0.26

(a). We need to calculate the friction force

Using formula of friction force

$f_{k}=\mu F_{N}$

Put the value into the formula

$f_{k}=0.26\times7.50\times10^{3}$

$f_{k}=1950\ N$

(b). If the only other horizontal force exerted on the sleigh is due to the horses pulling the sleigh,

We need to calculate the magnitude of this force

According to given data,

The same force will be applied to keep constant velocity.

Hence, (a). The kinetic friction force is 1950 N.

(b). The magnitude of force will be equal of friction force.

Answer 2

(a). The kinetic friction force is 1950 N.

(b). The magnitude of force will be equal of friction force

• The calculation is as follows;

a. The magnitude of the kinetic friction force experienced by the sleigh is

$= 0.76 \times 7.50 \times 10^3$

= 1950 N

b. It should be equivalent to the friction force.

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