We can calculate the work done by friction by using \(\displaystyle W=F_{friction}\cdot d\).

It is just the product of the friction force and the length of the ramp. Rewrite work as \(\displaystyle W=\mu_kF_Nd\).

Since this problem involves a box sliding down an incline, the normal force is \(\displaystyle F_N=mg\cos\theta\).

Work is now written as the equation below.

\(\displaystyle W=\mu_kmg(\cos\theta) d\)

We know the following information from the question.

\(\displaystyle \mu_k=0.4\)

\(\displaystyle m=287kg\)

\(\displaystyle g=9.8\frac{m}{s^2}\)

\(\displaystyle \theta=33^{o}\)

\(\displaystyle d=5.5m\)

using these values, we can calculate work from our equation.

\(\displaystyle W=(0.4)(287kg)(9.8\frac{m}{s^2})(\cos(33))(5.5)\)

\(\displaystyle W=5189 J\)

First, we know that the tension on the string is \(\displaystyle 40\:N\) at an angle of \(\displaystyle 30\) degrees from the horizontal. We must find the x and y components of the tension force.

\(\displaystyle T_y = T\sin(30) = 40\cdot\sin(30)= 20\:N\)

\(\displaystyle T_x = T\cos(30) = 40\cos(30)= 34.64\:N\)

Next, we must find the normal force acting on the block. The normal force of the block itself is \(\displaystyle 15\:kg \cdot 9.8\:m/s^2 = 147\:N\), but there is a string pulling on it. We must subtract the y component of tension to get the final normal force

\(\displaystyle F_n = 147-20 = 127\)

Next, we know that the block is being pulled to the right, and the frictional force is acting in the opposite direction. So we get the equation:

\(\displaystyle T_x - F_f = 0, F_f = \mu F_n\)

\(\displaystyle 34.64 - \mu 127 = 0\)

\(\displaystyle \mu = 0.273\)

\(\displaystyle \begin{align*}&\text{To solve a problem of force balance, a good idea is to draw}\\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\\&\text{of forces acting on it must be equal to zero. The forces acting}\\&\text{on it are gravity (in the form of the mass times the gravitational}\\&\text{constant), a normal force, and since it’s on an incline, a frictional}\\&\text{force to balance it all out:}\\&F_{mass}=m*g\\&F_{normal}=m*g*cos(a)\\&F_{friction}=m*g*sin(a))\\&F_{friction}=51kg(9.81\frac{m}{s^2})(sin(11^{\circ})\\&F_{friction}=95.46N\end{align*}\)

\(\displaystyle \begin{align*}&\text{To solve a problem of force balance, a good idea is to draw}\\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\\&\text{of forces acting on it must be equal to zero. The forces acting}\\&\text{on it are gravity (in the form of the mass times the gravitational}\\&\text{constant), a normal force, and since it’s on an incline, a frictional}\\&\text{force to balance it all out:}\\&F_{mass}=m*g\\&F_{normal}=m*g*cos(a)\\&F_{friction}=m*g*sin(a)\\&F_{friction}=100kg(9.81\frac{m}{s^2})(sin(73^{\circ}))\\&F_{friction}=938.13N\end{align*}\)

\(\displaystyle \begin{align*}&\text{To solve a problem of force balance, a good idea is to draw}\\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\\&\text{of forces acting on it must be equal to zero. The forces acting}\\&\text{on it are gravity (in the form of the mass times the gravitational}\\&\text{constant), a normal force, and since it’s on an incline, a frictional}\\&\text{force to balance it all out:}\\&F_{mass}=m*g\\&F_{normal}=m*g*cos(a)\\&F_{friction}=m*g*sin(a)\\&F_{friction}=19kg(9.81\frac{m}{s^2})(sin(69^{\circ}))\\&F_{friction}=174.01N\end{align*}\)

\(\displaystyle \begin{align*}&\text{To solve a problem of force balance, a good idea is to draw}\\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\\&\text{of forces acting on it must be equal to zero. The forces acting}\\&\text{on it are gravity (in the form of the mass times the gravitational}\\&\text{constant), the force of the rope supporting it, a normal force,}\\&\text{and since it’s on an incline, a frictional force to balance}\\&\text{it all out. Notice that the tension in the rope mitigates some}\\&\text{of the gravitational force (as illustrated in the upper left}\\&\text{corner) to lessen what is applied to the slope:}\\&F_{mass}=m*g\\&F_{normal}=(F_{mass}-F_{rope})*cos(a)\\&F_{friction}=(F_{mass}-F_{rope})*sin(a)\\&F_{rope}=F_{mass}-\frac{F_{friction}}{sin(a)}\\&F_{rope}=2786.04N-\frac{207.04N}{sin(48^{\circ})}\\&F_{rope}=2507.44N\end{align*}\)

\(\displaystyle \begin{align*}&\text{To solve a problem of force balance, a good idea is to draw}\\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\\&\text{of forces acting on it must be equal to zero. The forces acting}\\&\text{on it are gravity (in the form of the mass times the gravitational}\\&\text{constant), the force of the rope supporting it, a normal force,}\\&\text{and since it’s on an incline, a frictional force to balance}\\&\text{it all out. Notice that the tension in the rope mitigates some}\\&\text{of the gravitational force (as illustrated in the upper left}\\&\text{corner) to lessen what is applied to the slope:}\\&F_{mass}=m*g\\&F_{normal}=(F_{mass}-F_{rope})*cos(a)\\&F_{friction}=(F_{mass}-F_{rope})*sin(a)\\&F_{friction}=(101kg(9.81\frac{m}{s^2})-693.57N)(sin(68^{\circ}))\\&F_{friction}=275.6N\end{align*}\)

\(\displaystyle \begin{align*}&\text{To solve a problem of force balance, a good idea is to draw}\\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\\&\text{of forces acting on it must be equal to zero. The forces acting}\\&\text{on it are gravity (in the form of the mass times the gravitational}\\&\text{constant), a normal force, and since it’s on an incline, a frictional}\\&\text{force to balance it all out:}\\&F_{mass}=m*g\\&F_{normal}=m*g*cos(a)\\&F_{friction}=m*g*sin(a)\\&F_{friction}=187kg(9.81\frac{m}{s^2})(sin(70^{\circ}))\\&F_{friction}=1723.84N\end{align*}\)

\(\displaystyle \begin{align*}&\text{To solve a problem of force balance, a good idea is to draw}\\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\\&\text{of forces acting on it must be equal to zero. The forces acting}\\&\text{on it are gravity (in the form of the mass times the gravitational}\\&\text{constant), a normal force, and since it’s on an incline, a frictional}\\&\text{force to balance it all out:}\\&F_{mass}=m*g\\&F_{normal}=m*g*cos(a)\\&F_{friction}=m*g*sin(a)\\&F_{friction}=208kg(9.81\frac{m}{s^2})(sin(28^{\circ}))\\&F_{friction}=957.95N\end{align*}\)