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Answer to Question #113438 in Mechanics | Relativity for Emma

Question #113438
The 8-kg ball is connected with a spring as shown. Initially the spring is compressed by 0.2 m. Then under the applied force F, the spring is stretched to 0.6 m. Determine the total work done to the ball during this process. The slope is smooth. 300 Nm (a) 151 J (b) 254 J (c) 99.1 J (d) 306 J F=400 N 150 40°
1
2020-05-04T12:47:17-0400

There is work done by three different forces

(a) work of the force "F"

"W_F=F\\cdot s\\cdot \\cos35\u00b0=400\\cdot (0.2+0.6)\\cdot\\cos35\u00b0=262J"

(b) work of the block weight

"W_W=mg\\Delta h=mg\\cdot0.8\\cdot \\sin40\u00b0=8\\cdot 9.8\\cdot0.8\\cdot \\sin40\u00b0=40J"

(c) work of the spring force

"W_S=-\\frac{k}{2}(0.6^2-0.2^2)=-\\frac{300}{2}(0.36-0.04)=-48J"

So, total work

"W_t=262+40-48=254J"

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Assignment Expert
02.11.21, 16:12

Dear eve, 0.2+0.6=0.8 m

eve
02.11.21, 06:18

How did you come up with the value 0.8?