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Chapter 6: Problem 13

Sam pushes a \(10.0-\mathrm{kg}\) sack of bread flour on a frictionless horizontal surface with a constant horizontal force of \(2.0 \mathrm{N}\) starting from rest. (a) What is the kinetic energy of the sack after Sam has pushed it a distance of \(35 \mathrm{cm} ?\) (b) What is the speed of the sack after Sam has pushed it a distance of \(35 \mathrm{cm} ?\)

Short Answer

Expert verified
Question: A sack of bread flour with a mass of 10 kg is on a frictionless surface. A constant force of 2.0 N is used to push the sack a distance of 35 cm. Calculate (a) the kinetic energy of the sack after being pushed, and (b) the speed of the sack after being pushed. Answer: (a) The kinetic energy of the sack after being pushed for a distance of 35 cm is 0.7 J. (b) The speed of the sack after being pushed a distance of 35 cm is 0.37 m/s.

Step by step solution

01

Part (a): Kinetic Energy Calculation

We are given the initial kinetic energy \(K_{initial} = 0 \mathrm{J}\), the force applied \(F = 2.0 \mathrm{N}\) and the distance traveled \(d = 35 \mathrm{cm} = 0.35 \mathrm{m}\). We will use the work-energy principle, which states that the work done on an object is equal to its change in kinetic energy. The work done by the constant force can be calculated as follows: $$ W = F \cdot d $$ Once we find the work done, we can use it to find the final kinetic energy \(K_{final}\). Since the initial kinetic energy is zero, the final kinetic energy will be equal to the work done. $$ K_{final} = K_{initial} + W $$
02

Part (a): Calculate Work Done

Now we can calculate the work done using the given force and distance: $$ W = F \cdot d = (2.0 \mathrm{N}) \cdot (0.35 \mathrm{m}) = 0.7\ \mathrm{J} $$
03

Part (a): Calculate Final Kinetic Energy

Since the initial kinetic energy is zero, the final kinetic energy will be equal to the work done: $$ K_{final} = K_{initial} + W = 0 + 0.7\ \mathrm{J} = 0.7\ \mathrm{J} $$ The kinetic energy of the sack after being pushed for a distance of \(35 \mathrm{cm}\) is \(0.7\ \mathrm{J}\).
04

Part (b): Speed Calculation

Now we need to find the speed of the sack after being pushed for \(35\ \mathrm{cm}\). The kinetic energy (\(K\)) of an object is related to its mass (\(m\)) and its speed (\(v\)) by the following equation: $$ K = \frac{1}{2}mv^2 $$ Since we have found the final kinetic energy, we can use this equation to find the speed of the sack: $$ v = \sqrt{\frac{2K}{m}} $$
05

Part (b): Calculate Speed of the Sack

We will now calculate the speed of the sack using the final kinetic energy and its mass: $$ v = \sqrt{\frac{2 \cdot 0.7\ \mathrm{J}}{10.0\ \mathrm{kg}}} = \sqrt{\frac{1.4\ \mathrm{J}}{10.0\ \mathrm{kg}}} = \sqrt{0.14\ \mathrm{m^2/s^2}} = 0.37\ \mathrm{m/s} $$ The speed of the sack after being pushed a distance of \(35\ \mathrm{cm}\) is \(0.37\ \mathrm{m/s}\).

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