A 0.500 kg block is being slid, vertically up the chalkboard with an applied force of 6.00 N [upwards] and 2.00 N [inwards] (into the boards). If the coefficient of sliding friction is 0.400, calculate the accel of the block.
net force=ma
6-mg-2*.4=ma
solve for a
can i get a better explanation since im having a hard time getting the concept. thanks =)
To calculate the acceleration of the block, we need to consider the forces acting on it. The forces acting on the block are the gravitational force (mg), the applied force (F_applied), and the force of sliding friction (F_friction).
First, let's calculate the gravitational force acting on the block:
F_gravity = mg
Given:
mass (m) = 0.500 kg
acceleration due to gravity (g) = 9.8 m/s^2
F_gravity = (0.500 kg) * (9.8 m/s^2)
F_gravity = 4.9 N
Next, calculate the net force acting on the block:
Net Force (F_net) = F_applied + F_friction
The applied force (F_applied) is given as 6.00 N upward and 2.00 N inward. Since the block is moving vertically up, only the upward component will contribute to the net force. So, F_applied = 6.00 N.
The force of sliding friction (F_friction) can be calculated using the equation:
F_friction = coefficient of sliding friction * normal force
The normal force (F_normal) is the force exerted by the surface on the block perpendicular to it. In this case, the normal force is equal to the gravitational force (F_gravity). So, F_normal = F_gravity = 4.9 N.
Now, substitute the given coefficient of sliding friction (μ = 0.400) and normal force (F_normal = 4.9 N) into the equation to calculate the force of sliding friction:
F_friction = (0.400) * (4.9 N)
F_friction = 1.96 N
Finally, calculate the net force:
F_net = F_applied + F_friction
F_net = 6.00 N + 1.96 N
F_net = 7.96 N
Now, use Newton's second law of motion:
F_net = ma
Rearranging the equation, we get:
a = F_net / m
Substituting the known values:
a = 7.96 N / 0.500 kg
a = 15.92 m/s^2
Therefore, the acceleration of the block is 15.92 m/s^2.