15kg block rests on a rough surface (u=.35) a 100N force to the left is applied to the block.
what is the magnitude and direction of the net force acting on the block?
How quickly will the block accelerate over the rough surface?
1. Fb = mg = 15kg * 9.8N/kg = 147N.
147N @ 0 deg.
Fp = 147sin(0) = 0 = Force parallel to plane(hor.).
Fv = 147cos(0) = 147N = Force perpendicular to plane.
Ff = u*Fv = 0.35 * 147 = 51.45N.
Fn = Fap - Fp - Ff,
Fn = 100 - 0 - 51.45 = 48.6N = Net force parallel to the hor plane.
To determine the magnitude and direction of the net force acting on the block, we need to consider the forces acting on the block.
1. Weight force (F_w):
The weight force of the block can be calculated using the formula F_w = m * g, where m is the mass and g is the acceleration due to gravity. Since the mass is 15 kg and the acceleration due to gravity is approximately 9.8 m/s^2, the weight force is:
F_w = 15 kg * 9.8 m/s^2 = 147 N (directed downwards)
2. Applied force (F_applied):
An external force of 100N is applied to the block, directed to the left.
3. Friction force (F_friction):
The friction force can be calculated using the formula F_friction = u * F_n, where u is the coefficient of friction and F_n is the normal force. The normal force is equal to the weight force in this case.
F_friction = 0.35 * 147 N = 51.45 N (direction opposite to the applied force)
Now, let's calculate the net force:
Net Force = F_applied + F_friction
Net Force = 100 N + (-51.45 N) = 48.55 N
The magnitude of the net force is 48.55 N, and the direction is to the left.
To determine the acceleration of the block over the rough surface, we can use Newton's second law of motion. The net force is equal to the mass multiplied by the acceleration (F_net = m * a).
Rearranging the formula, we have:
a = F_net / m
Substituting the values:
a = 48.55 N / 15 kg = 3.237 m/s^2
The block will accelerate over the rough surface at a rate of 3.237 m/s^2.