A winch is used to frag a 375 N crate up a ramp at a constant speed of 0.75 m/s by means of a rope that pulls parallel to surface of the ramp. The rope slopes at 33 deg. above the horizontal, and the coefficient of kinetic friction between the ramp and the crate is 0.25. (a) What is the tension in the rope? (b) If the rope were suddenly to snap, what would be the acceleration of the crate immediately after the rope broke?

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a) The tension in the rope can be calculated using Newton's second law in vertical (y) and horizontal (x) directions.

The forces acting on the crate are:

- Gravity (mg) acting vertically downward
- Tension (T) in the rope acting 33 degrees above the horizontal
- Normal force (N) acting perpendicular to the ramp
- Friction force (f) acting parallel to the ramp

First, we solve for the vertical components of the forces:

T_y = T*sin(33) (component of the tension in the y direction)
N_y = N (since the normal force is perpendicular to the ramp)
f_y = 0 (since friction is parallel to the ramp)
g_y = -375 * 9.81 * sin(33) (component of the gravitational force in the y direction)

To get the net force in the y direction, we sum these components:
F_net_y = T_y + N_y + f_y + g_y

The net force in the vertical direction is zero, and there is no acceleration in the vertical direction, so we can set up the following equation:
0 = T*sin(33) + N - 375 * 9.81 * sin(33)

Now we solve for the horizontal components of the forces:

T_x = T*cos(33) (component of the tension in the x direction)
N_x = 0 (since the normal force is perpendicular to the ramp)
f_x = -0.25 * N (since friction is parallel to the ramp and opposite to the motion)
g_x = -375 * 9.81 * cos(33) (component of the gravitational force in the x direction)

To get the net force in the x direction, we sum these components:
F_net_x = T_x + N_x + f_x + g_x

Since the crate is moving up the ramp at a constant speed, the net force in the x direction is also zero. Therefore, we can set up the following equation:
0 = T*cos(33) - 0.25 * N - 375 * 9.81 * cos(33)

We now have a system of two equations with two unknowns:

(1) 0 = T*sin(33) + N - 375 * 9.81 * sin(33)
(2) 0 = T*cos(33) - 0.25 * N - 375 * 9.81 * cos(33)

We will solve the system using the method of substitution. We can rearrange equation (1) to solve for N:

N = T*sin(33) + 375 * 9.81 * sin(33)

Now we substitute this expression for N into equation (2):

0 = T*cos(33) - 0.25 * (T*sin(33) + 375 * 9.81 * sin(33)) - 375 * 9.81 * cos(33)

Now we can solve for T:

T = (375 * 9.81 * cos(33) + 0.25 * 375 * 9.81 * sin(33))/(cos(33) - 0.25 * sin(33))

T ≈ 844.61 N

So, the tension in the rope is approximately 844.61 N.

b) After the rope snaps, there are three forces acting on the crate: the gravitational force, the normal force, and the friction force. We will find the acceleration in the horizontal (x) direction, which is parallel to the ramp.

First, we need to update the friction force since it is proportional to the normal force. Using the expression for N we found earlier, we can find the normal force after the rope snaps:

N_snap = 375 * 9.81 * sin(33)
N_snap ≈ 1992.65 N

Now we can find the friction force after the rope snaps:

f_x_snap = -0.25 * N_snap
f_x_snap ≈ -498.16 N

We can now find the net force in the horizontal direction after the rope snaps:

F_net_x_snap = f_x_snap + g_x
F_net_x_snap ≈ -498.16 - 375 * 9.81 * cos(33)
F_net_x_snap ≈ -1663.95 N

Now we can use Newton's second law to find the acceleration in the horizontal direction:

a_x_snap = F_net_x_snap / 375
a_x_snap ≈ -4.44 m/s²

So, if the rope were to snap, the acceleration of the crate immediately after would be approximately -4.44 m/s².