Question

2. a 35 kg wagon is being pushed with a metal rod that makes an angle of 48o with the horizontal. the applied force is 449 n and the coefficient of kinetic friction between the wagon and the ground is 0.19.
I'm not sure what equation to use to find the normal force and then the acceleration

Answer 1

The acceleration required to push the wagon is 16.12m/s².

Acceleration is the rate of change of velocity with respect to time.

• Also it defines the maximum capacity of a vehicle to attain its speed.
• Its SI unit is m/sec².

According to Newtons Second law of motion,

Force = Mass * acceleration

Fnet = m*a

Fnet = Fm + Ff

Fm = Force due to Mass of the wagon = mg = 449N

Ff = Force due to friction = μR

Fnet = mg + μR

μ = 0.19

m = 35kg

R = mg; g = 9.8 m/s²

Fm + Ff = ma

449 + (0.19*35*9.8) = 35*a

499 + 65.17 = 35a

564.17 = 35a

a = 16.12 m/s²

The acceleration required to push the wagon is 16.12m/s² and the normal force will be the force acting due to weight of the wagon i.e. 449N.

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