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A skater pushes a 3.77 kg skateboard with a 6.23 N force directed at 47.4 degrees above horizontal. A resistance force, due to friction, of 4.50 N opposes the motion of the skateboard. What is the acceleration of the skateboard? Remember to indicate if the acceleration is positive (+) or negative (-).

Respuesta :

Answer:

Step 1: Resolve the force applied into horizontal and vertical components.

Fhorizontal F. cos(0)

Fhorizontal 6.23 N. cos(47.4°)

Fhorizontal 4.20 N

Step 2: Calculate the net force acting horizontally.

Fnet Fhorizontal - Ffriction

Fnet 4.20 N - 4.50 N

Fnet -0.30 N

Step 3: Use Newton's second law to find the acceleration.

Fnet = ma

-0.30 N 3.77 kg a

-0.30 N

a 3.77 kg

a≈ -0.079 m/s²

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