4. We use the results from section 11.3. (a) We substitute IMR=252(Table 10-2(f)) and a= – 0.10ginto Eq. 11-10: −=−+=−0101752522.sin/ggMRMRgθchwhich yields = sin–1(0.14) = 8.0°. (b) The acceleration would be more. We can look at this in terms of forces or in terms of energy. In terms of forces, the uphill static friction would then be absent so the downhill acceleration would be due only to the downhill gravitational pull. In terms of energy, the
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.