Question

a concrete block is pulled 7.0 m across a frictionless surface by means of a rope. the tension in the rope is 40 n; and the net - work done on the block is 247 j. what angle does the rope make with the horizontal? the kinetic energy of a car is 8×10^8 j as it travels along a horizontal road. how much work is required to stop the car in 10 s?

Explanation:

Step1: Recall work - formula

The work - done formula is $W = Fd\cos\theta$, where $W$ is the work done, $F$ is the force, $d$ is the displacement, and $\theta$ is the angle between the force and the displacement.
We are given that $W = 247\ J$, $F = 40\ N$, and $d = 7.0\ m$.
Substituting these values into the formula, we get $247=40\times7\times\cos\theta$.

Step2: Solve for $\cos\theta$

First, simplify the right - hand side of the equation: $40\times7 = 280$. So, the equation becomes $247 = 280\cos\theta$.
Then, solve for $\cos\theta$: $\cos\theta=\frac{247}{280}\approx0.8821$.

Step3: Find the angle $\theta$

Take the inverse cosine of $\cos\theta$: $\theta=\cos^{- 1}(0.8821)$.
Using a calculator, $\theta\approx28.1^{\circ}$.

For the second part:

Step1: Recall the work - energy theorem

The work - energy theorem states that the net work done on an object is equal to the change in its kinetic energy, $W=\Delta K$.
The initial kinetic energy of the car is $K_i = 8\times10^{8}\ J$, and the final kinetic energy of the car when it stops is $K_f = 0\ J$.

Step2: Calculate the work required

The change in kinetic energy $\Delta K=K_f - K_i=0 - 8\times10^{8}\ J=- 8\times10^{8}\ J$.
The work required to stop the car is equal to the change in its kinetic energy. So, $W=-8\times10^{8}\ J$. The negative sign indicates that the work is done against the motion of the car.

Answer:

The angle the rope makes with the horizontal is approximately $28.1^{\circ}$. The work required to stop the car is $8\times10^{8}\ J$ (the magnitude of the work).