# An object with a mass of #6 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= 2+cscx #. How much work would it take to move the object over #x in [pi/8, (7pi)/8], where x is in meters?

The work is

We need

The work done is

The frictional force is

The work done is

By signing up, you agree to our Terms of Service and Privacy Policy

To calculate the work done to move the object over the given interval [π/8, (7π)/8], we need to integrate the force of kinetic friction over this interval. The formula for work done against friction is:

[ W = \int_{x_1}^{x_2} F_{\text{friction}} , dx ]

Given that the kinetic friction coefficient ( u_k(x) = 2 + \csc(x) ), we can find the force of kinetic friction ( F_{\text{friction}} ) using the formula:

[ F_{\text{friction}} = u_k(x) \times \text{normal force} ]

The normal force equals the weight of the object, which is ( mg ), where ( m ) is the mass of the object and ( g ) is the acceleration due to gravity (( g \approx 9.8 , \text{m/s}^2 )).

So, we have:

[ F_{\text{friction}} = (2 + \csc(x)) \times mg ]

Substituting the given mass (( m = 6 ) kg), and integrating over the interval [π/8, (7π)/8], we get:

[ W = \int_{\frac{\pi}{8}}^{\frac{7\pi}{8}} (2 + \csc(x)) \times 6 \times 9.8 , dx ]

[ W = 6 \times 9.8 \int_{\frac{\pi}{8}}^{\frac{7\pi}{8}} (2 + \csc(x)) , dx ]

[ W = 6 \times 9.8 \left[ 2x - \ln|\csc(x) + \cot(x)| \right]_{\frac{\pi}{8}}^{\frac{7\pi}{8}} ]

[ W \approx 6 \times 9.8 \left[ \left( 2\frac{7\pi}{8} - \ln|\csc\left(\frac{7\pi}{8}\right) + \cot\left(\frac{7\pi}{8}\right)| \right) - \left( 2\frac{\pi}{8} - \ln|\csc\left(\frac{\pi}{8}\right) + \cot\left(\frac{\pi}{8}\right)| \right) \right] ]

[ W \approx 6 \times 9.8 \left[ \left( \frac{7\pi}{4} - \ln|\csc\left(\frac{7\pi}{8}\right) + \cot\left(\frac{7\pi}{8}\right)| \right) - \left( \frac{\pi}{4} - \ln|\csc\left(\frac{\pi}{8}\right) + \cot\left(\frac{\pi}{8}\right)| \right) \right] ]

[ W \approx 6 \times 9.8 \left[ \left( \frac{7\pi}{4} - \ln\left|\csc\left(\frac{7\pi}{8}\right) + \cot\left(\frac{7\pi}{8}\right)\right| \right) - \left( \frac{\pi}{4} - \ln\left|\csc\left(\frac{\pi}{8}\right) + \cot\left(\frac{\pi}{8}\right)\right| \right) \right] ]

[ W \approx 6 \times 9.8 \left( \frac{7\pi}{4} - \ln\left|\csc\left(\frac{7\pi}{8}\right) + \cot\left(\frac{7\pi}{8}\right)\right| - \frac{\pi}{4} + \ln\left|\csc\left(\frac{\pi}{8}\right) + \cot\left(\frac{\pi}{8}\right)\right| \right) ]

By signing up, you agree to our Terms of Service and Privacy Policy

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

- An object with a mass of #5 kg# is hanging from a spring with a constant of #3 (kg)/s^2#. If the spring is stretched by #2 m#, what is the net force on the object?
- How can frictional forces be reduced?
- How fast will an object with a mass of #3 kg# accelerate if a force of #7 N# is constantly applied to it?
- How can inertia be measured?
- In a pulley system, #m_1# is #2.0kg# and #m_2# is #1.0kg#. The coefficient of kinetic friction between #m_1# and the table is #mu_k=0.15#. What is the acceleration of the pulley system?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7