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Chapter 11: Problem 6

A \(47.2-\mathrm{kg}\) block of ice slides down an incline \(1.62 \mathrm{~m}\) long and \(0.902 \mathrm{~m}\) high. A worker pushes up on the ice parallel to the incline so that it slides down at constant speed. The coefficient of kinetic friction between the ice and the incline is \(0.110 .\) Find \((a)\) the force exerted by the worker, \((b)\) the work done by the worker on the block of ice, and \((c)\) the work done by gravity on the ice.

Short Answer

Expert verified
The force exerted by the worker and the work done by gravity will be positive values since they are in the direction of displacement, while the work done by the worker will have a negative sign as it is against the direction of displacement. Exact values will be obtained after substituting the given data into the derived expressions.

Step by step solution

01

Calculate the angle of inclination

Considering the right triangle formed by the incline, the angle of inclination \( \theta \) can be calculated by tan inverse of the ratio of height to length. Hence, using the formula \( \theta = \tan^{-1}(\frac{0.902}{1.62}) \)
02

Calculate the force exerted by the worker

The block of ice slides down at constant speed. This means that the net force along the incline is zero. Therefore, the force exerted by the worker must balance the gravitational force and the frictional force. The force exerted by the worker is given by \( F_\text{worker} = F_\text{gravity} + F_\text{friction} = m \cdot g \cdot \sin\theta + \mu \cdot m \cdot g \cdot \cos\theta \) where m is the mass of the block, g is the acceleration due to gravity, \( \mu \) is the coefficient of friction, and \( \theta \) is the angle of the incline.
03

Calculate the work done by the worker and gravity

The work done by a force during displacement is given by \( W = F \cdot s \cdot \cos\alpha \) where F is the magnitude of the force, s is the displacement and \( \alpha \) is the angle between the force and the displacement. So, work done by the worker is \( W_\text{worker} = F_\text{worker} \cdot s \cdot \cos180^o = - F_\text{worker} \cdot s \) and the work done by gravity is \( W_\text{gravity}= m \cdot g \cdot h \)

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