Question

You throw a 20 N rock into the air from ground level and observe that, when it is 15.0 m high, it is traveling upward at 25.0 m/s. Use the work-energy principle to finda. the rock's speed just as it left the ground andb. the maximum height the rock will reach

Answer 1

Answer:

a) 30.3 m/s b) 46.8 m

Explanation:

a)

• Assuming no friction present, the work-energy theorem says that the work done on the rock by the only force acting on it (gravity force, aiming downward), is equal to the change of kinetic energy of the rock, at any point.
• When it's at 15.0 m high, the work done by gravity on the rock can be written as follows:

        $W = m*g* h* cos 180\º = -20 N* 15.0 m = -300 J (1)$

• The change in kinetic energy of the rock can be expressed in this way:

        $\Delta K = K_{f} - K_{0} = \frac{1}{2} * m* (v_{f}^{2} -v_{0} ^{2} ) (2)$

• where m= 2.04 kg, vf = 25.0 m/s.
• As (1) and (2) are equal each other, we can solve for v₀, as follows:

        $= -300 J = 0.5*2.04 kg *(( 25.0 m/s)^{2} -v_{0}^{2} )\\ \\ v_{0}^{2} = \frac{300*2 J}{2.04kg} + 625 (m/s)2 = 894. 1 (m/s)2 \\ \\ v_{0} = \sqrt{894.1 (m/s)2} = 30.3 m/s$

b)

• When the rock arrives to its highest point, it will momentarily come to a stop, before changing direction to start falling.
• So, the final kinetic energy, will be zero.
• Applying the work --energy theorem again, we can write the following equation:

        $\Delta K = K_{f} - K_{0} =( 0- \frac{1}{2} *m*v_{0}^{2}) = -m*g*h_{max}$

• Replacing by v₀ = 30.3 m/s, rearranging and simplifying common terms, we can solve for hmax, as follows:

        $h_{max} = \frac{v_{0} ^{2} }{2*g} =\frac{(30.3 m/s)^{2}}{2*9.8 m/s2} = 46.8 m$