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Question 4
Figure 10 shows a toy used to launch a ball. One end of the spring is fixed to the handle. The other end of the spring is fixed to the support. (a) A child pulls t... show full transcript
Step 1
Answer
To determine the correct diagram illustrating the forces acting on the handle, we need to consider the tension force exerted by the stretched spring. The force diagram should include the force acting horizontally towards the right (the force exerted by the spring) and the force acting to the left (the reactive force at the handle). The correct option will effectively show these forces.
Step 2
Step 3
Answer
To calculate the work done in extending the spring, we use the formula:
ext{work done} = rac{1}{2} imes k imes ( ext{extension})^2Substituting in the given values:
ext{work done} = rac{1}{2} imes 20 ext{ N/m} imes (0.09 ext{ m})^2Calculating this gives:
ext{work done} = rac{1}{2} imes 20 imes 0.0081 = 0.081 ext{ J}Therefore, the work done in extending the spring by 0.09 m is 0.081 J.
Step 4
Answer
When the child releases the handle, the potential energy stored in the stretched spring is converted into kinetic energy. Initially, the energy in the spring is potential energy, and as the spring returns to its original length, this energy is transferred to the ball, converting into its kinetic energy, causing the ball to accelerate and move forward.
Step 5
Answer
The design of the toy ensures that the spring is only stretched until the pad is pressed against the support. This limits the extension of the spring, preventing it from exceeding its maximum elastic limit. By controlling the maximum extension, the toy prevents overstretching that could lead to permanent deformation or breakage of the spring, thus maintaining its functionality and longevity.
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