Question

1. the work done on a slingshot is 40.0 joules to pull back a 0.10 - kilogram stone. if the slingshot projects the stone straight up in the air, what is the maximum height to which the stone will rise? neglect friction. (1) 0.41 m (2) 41 m (3) 410 m (4) 4.1 m

pe
w = 40j
m = 0.10 kg
δh =?
pe = ke
δpe = mgδh
(40j) = (0.10kg)(9.81m/s²)δh
40j = 0.981δh
(40j)/0.981 = (0.981δh)/0.981
41m = δh

Explanation:

Step1: Identify Energy Conversion

The work done on the slingshot (40.0 J) is converted into kinetic energy (KE), which then converts to gravitational potential energy (PE) at maximum height. So, \( PE = KE = 40.0 \, \text{J} \).

Step2: Recall Potential Energy Formula

The formula for gravitational potential energy change is \( \Delta PE = mg\Delta h \), where \( m = 0.10 \, \text{kg} \), \( g = 9.81 \, \text{m/s}^2 \), and \( \Delta h \) is the height change.

Step3: Solve for \( \Delta h \)

Rearrange the formula: \( \Delta h = \frac{\Delta PE}{mg} \). Substitute \( \Delta PE = 40.0 \, \text{J} \), \( m = 0.10 \, \text{kg} \), \( g = 9.81 \, \text{m/s}^2 \):
\( \Delta h = \frac{40.0}{(0.10)(9.81)} = \frac{40.0}{0.981} \approx 40.8 \, \text{m} \), which rounds to 41 m.

Answer:

41 m (corresponding to the option marked, likely option 2 if numbered as (1) 0.41 m, (2) 41 m, (3) 410 m, (4) 4.1 m)