A father goes out with his young child in a sled. The child plus sled has a mass

Question

A father goes out with his young child in a sled. The child plus sled has a mass of 65 kg. The sled has a length of 2.2 m. The father pulls the first sled with a force of 250 N in a direction of 30 degrees above the horizontal. The frictional coefficient of the snow is 0.045. The sled moves only horizontally with an initial velocity of 1.2 m/s forward. What is the net force on the sled in the Vertical direction? In Newtons how large is the normal force? What is the net force on the sled in the HORIZONTAL direction? What is the acceleration of the sled?

Explanation / Answer

Mass child +sled=>M=65 kg

The sled has length L=2.3 m

Applied force F=250 N

Angle =30°

Co efficient of friction =0.045

Initial velocity u=1.2 m/s

This concept belongs to laws of motion

According to the friction

(a)

The net force in the vertical direction

N=mg cos(theta)

=65×9.8×cos30°

=551.7

But the net force =mg+mgcos(theta)

=65×9.8+551.7

=637+551.7=1188.7 N

(B) and

The normal force N=mg cos(30°)

=551.7N

(C) and

The net force applied on the sled in horizontal direction

Fnet=F+mg{(sin(theta)+(co.efficient of friction) cos(theta)}

=250+65×9.8{(sin(30)+0.045cos(30)}

=593.3 N

(D) ans

Acceleration =F/m=250/65=3.85m/s^2

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