A 450 g block is at rest on a horizontal surface. The coefficient of static friction between the block and the surface is 0.670, and the coefficient of kinetic friction is 0.380. A force of magnitude P pushes the block forward and downward as shown in the figure below. Assume the force is applied at an angle of 39.0° below the horizontal.
(a) Assuming P is large enough to make the block move, find the acceleration of the block as a function of P. (Use P as necessary and round numbers to the third decimal place.) (b) If P = 6.81 N, find the acceleration and the friction force exerted on the block.
a = m/s2
Ffriction = N
(c) If P = 13.62 N, find the acceleration and the friction force exerted on the block.
a = m/s2
Ffriction = N
(d) Describe in words how the acceleration depends on P.
Wb = mg = 0.45kg * 9.8N/kg = 4.41N. =
Weight of block.
Fb = (4.41N,0deg.).
Fp=Fh = 4.41sin(0) = 0 = Force parallel
to hor. surface.
Fv = 4.41cos(0) = 4.41N. = Force perpendicular to hor. surface.
Ff = u*Fv = 0.67 * 4.41 = 2.95N. = Force of static friction.
Ff = u*Fv = 0.360 * 4.41 = 1.59N. = Force of kinetic friction.
a. P must overcome the static friction to move the block:
Pcos39 - Ff = 0,
Pcos39 = Ff,
Pcos39 = 2.95,
P = 2.95 / cos39 = 3.8N. = Force applied.
When block starts to move, P must overcome kinetic friction to accelerate:
Fn = Pcos39 - Ff = 2.95 - 1.59 = 1.36 = Net force.
a = Fn / m = 1.36 / 0.45 = 3.03m/s^2.
b. Fn = 6.81cos39 - 1.59 = 3.7N.
a = Fn / m = 3.7 / 0.45 = 8.23m/s^2.
c. Fn = 13.62cos39 - 1.59 = 8.99N.
a = 8.99 / 0.45 = 19.98m/s^2.
d. Increasing P causes an increase in the net force, and the acceleration is proportional to the net force(a=Fn/m).
To find the acceleration of the block, we need to consider the forces acting on it. The forces in this case are the force P pushing the block forward and downward, the force of gravity pulling the block downward, and the frictional forces between the block and the surface.
(a) To find the acceleration as a function of P, we need to compare the forces acting on the block. The force pushing the block forward and downward can be divided into two components: one parallel to the surface and one perpendicular to the surface. The component parallel to the surface will help overcome friction and move the block.
The force of gravity pulling the block downward can be divided into two components as well: one perpendicular to the surface and one parallel to the surface. The component perpendicular to the surface balances out the normal force from the surface, and the component parallel to the surface contributes to the frictional force.
Since the block is at rest, the static frictional force exactly balances out the component of the force P parallel to the surface, meaning:
P_parallel = frictional force
The maximum static frictional force can be found using the equation:
f_static_max = μ_static * N
where μ_static is the coefficient of static friction and N is the normal force. The normal force is equal to the weight of the block, which can be calculated as:
N = m * g
where m is the mass of the block and g is the acceleration due to gravity.
Now, we can substitute these values into the equation for the maximum static frictional force:
f_static_max = μ_static * m * g
Since the block is just about to move, the static frictional force can be at its maximum value:
f_static = f_static_max
Therefore,
P_parallel = μ_static * m * g
The net force acting on the block is given by:
net force = P_parallel - frictional force
Using Newton's second law (F = m * a), we can equate the net force with the product of mass and acceleration:
P_parallel - frictional force = m * a
Now, we can solve for acceleration:
a = (P_parallel - frictional force) / m
Substituting the value for P_parallel and frictional force:
a = (μ_static * m * g - μ_static * m * g) / m
Simplifying further, we find that:
a = 0
This means that the block does not accelerate and remains at rest.
(b) When P = 6.81 N, the net force acting on the block is given by:
net force = P_parallel - frictional force
To find the acceleration, we can use the equation:
a = (P_parallel - frictional force) / m
The frictional force can be calculated using the equation:
frictional force = μ_kinetic * N
where μ_kinetic is the coefficient of kinetic friction.
The normal force can be calculated as:
N = m * g
where g is the acceleration due to gravity.
Substituting these values into the equation for the frictional force:
frictional force = μ_kinetic * m * g
Now, we can substitute the values for P_parallel and frictional force into the equation for acceleration:
a = (P_parallel - frictional force) / m
Solving for acceleration:
a = (P_parallel - μ_kinetic * m * g) / m
Substituting the given values for P, μ_kinetic, m, and g:
a = (6.81 * sin(39.0°) - 0.380 * 0.450 * 9.81) / 0.450
Calculating this value will give you the acceleration.
To find the friction force exerted on the block, we can substitute the acceleration and the given values into the equation:
frictional force = μ_kinetic * m * g
Calculating this value will give you the frictional force.
(c) Similar to part (b), when P = 13.62 N, you can follow the same steps to find the acceleration and the friction force exerted on the block.
(d) The acceleration of the block depends on P as follows:
- When P is small or equal to zero, the block does not accelerate and remains at rest.
- When P is large enough, just about any value greater than zero, the block will start to accelerate. The magnitude of acceleration depends on the value of P.