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kotykmax [81]
3 years ago
10

A 20 kg wagon is pulled along the level ground by a rope inclined at 30 degree above the horizontal. A friction force of 30 N op

poses the motion. How large is the pulling force if the wagon is moving with (a) constant speed and (b) and acceleration of 0.40 m/s2
Physics
1 answer:
Elan Coil [88]3 years ago
6 0

(a) 34.6 N

To solve the problem, we have to analyze the forces acting along the horizontal direction.

We have:

- Forward: the component of the pull parallel to the ground, which is

F cos \theta

where

F is the magnitude of the pull

\theta=30^{\circ} is the angle

- Backward: the force of friction, which is

F_f = 30 N

So, the equation of motion is

F cos \theta - F_f = ma

where

m = 20 kg is the mass of the wagon

a is the acceleration

In this part, the wagon is moving at constant speed, so a =0 and the equation becomes

F cos \theta - F_f = 0

Therefore, we can find the pulling force:

F=\frac{F_f}{cos \theta}=\frac{30}{cos 30}=34.6 N

(b) 43.9 N

In this case, the acceleration is

a=0.40 m/s^2

So, the equation of motion in this case is

F cos \theta - F_f = ma

So this time we have to take into account the term (ma).

Using the  same data as before:

m = 20 kg

\theta=30^{\circ}

F_f = 30 N

We find the new magnitude of F:

F=\frac{ma+F_f}{cos \theta}=\frac{(20)(0.40)+30}{cos 30}=43.9 N

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