A 3.90 kg block located on a horizontal frictionless floor is pulled by a cord that exerts a force F=14.1N at an angle theta=32.5degrees above the horizontal, as shown.
a.) What is the magnitude of the acceleration of the block when the force is applied?
b.) What is the horizontal speed of the block 4.50 seconds after it starts moving?
c.)What is the magnitude of the normal force acting on the block when the force F is acting on it?
d.)If, instead, the floor has a coefficient of kinetic friction µk = 0.04, what is the magnitude of the frictional force on the block when the block is moving?
e.)What is the magnitude of the acceleration of the block when friction is being considered?
I got all of the answers except e!
a.) 14.1cos32.5 / 3.90 = 3.05
b.) V=Vo+at
0+3.05(4.5)= 13.725
c.) fv=mg-fsin32.5
3.90(9.8)-14.1sin32.5 = 30.64
d.) 0.04(30.64)
=1.2256
Cant figure out e. please help!
e.) (14.1-1.23)cos32.5 / 3.90 = 2.78 m/s^2
Well, let's see if I can help you with that! To find the magnitude of the acceleration of the block when friction is being considered, we need to take into account the force of friction.
The force of friction can be calculated using the formula:
frictional force = coefficient of friction * normal force
In this case, the normal force is equal to the weight of the block, since the block is on a horizontal frictionless floor. So, the normal force can be calculated as:
normal force = mass * gravity
normal force = 3.90 kg * 9.8 m/s^2
Now, to find the magnitude of the acceleration, we can use Newton's second law:
acceleration = (net force) / mass
The net force acting on the block can be calculated by subtracting the force of friction from the force of the applied force:
net force = applied force - frictional force
And finally, substitute these values into the equation to find the magnitude of the acceleration:
acceleration = (applied force - frictional force) / mass
Plug in the values you know, and let's calculate the magnitude of the acceleration together!
To determine the magnitude of the acceleration of the block when friction is being considered, we need to consider the net force acting on the block.
The horizontal component of the applied force is given by F_hor = F * cos(theta):
F_hor = 14.1 * cos(32.5) = 11.913 N
The vertical component of the applied force is given by F_ver = F * sin(theta):
F_ver = 14.1 * sin(32.5) = 7.328 N
a.) Magnitude of the acceleration of the block when the force is applied:
The net force acting on the block in the horizontal direction is the component of the applied force in the horizontal direction, minus the force of friction:
F_net_hor = F_hor - f_friction
Since the floor is frictionless in this case, the force of friction is zero:
F_net_hor = F_hor = 11.913 N
The acceleration of the block is given by the equation F_net_hor = m * a:
11.913 = 3.90 * a
a = 3.06 m/s^2
Therefore, the magnitude of the acceleration of the block when the force is applied is 3.06 m/s^2.
e.) Magnitude of the acceleration of the block when friction is being considered:
Since there is now friction, we need to consider the force of friction in the horizontal direction. The force of friction is given by the equation f_friction = μ_k * N, where μ_k is the coefficient of kinetic friction and N is the normal force.
To find the normal force, we start by finding the vertical component of the weight of the block:
Weight_vertical = m * g * cos(theta) = 3.90 * 9.8 * cos(32.5) = 30.10 N
The normal force is equal to the weight of the block minus the vertical component of the applied force:
N = Weight - F_ver = 30.10 - 7.328 = 22.772 N
The force of friction is then given by:
f_friction = μ_k * N = 0.04 * 22.772 = 0.91088 N
The net force acting on the block in the horizontal direction is now:
F_net_hor = F_hor - f_friction = 11.913 - 0.91088 = 11.0021 N
Again, using the equation F_net_hor = m * a, we can find the acceleration:
11.0021 = 3.90 * a
a = 2.826 m/s^2
Therefore, the magnitude of the acceleration of the block when friction is being considered is 2.826 m/s^2.
To find the magnitude of the acceleration of the block when friction is considered, you need to calculate the net force acting on the block and then divide it by the mass of the block.
First, find the force of friction using the coefficient of kinetic friction and the normal force. The normal force can be found by subtracting the vertical component of the applied force F from the weight of the block.
c.) Normal force (Fn) = mg - Fsinθ
= 3.90 kg * 9.8 m/s^2 - 14.1 N * sin(32.5°)
≈ 30.64 N
d.) Frictional force (Fk) = μk * Fn
= 0.04 * 30.64 N
≈ 1.2256 N
Next, determine the net force acting horizontally on the block by adding the horizontal component of the applied force F and subtracting the frictional force Fk:
Net force (Fnet) = Fcosθ - Fk
= 14.1 N * cos(32.5°) - 1.2256 N
≈ 11.762 N
Finally, divide the net force by the mass to find the acceleration:
e.) Acceleration (a) = Fnet / m
= 11.762 N / 3.90 kg
≈ 3.01 m/s^2
Therefore, the magnitude of the acceleration of the block, taking into account friction, is approximately 3.01 m/s^2.