The coefficient of static friction between a 3.00 kg crate and the 35.0o incline is 0.300. What minimum force F must be applied perpendicularly to the incline to prevent the crate from sliding down

32.13 N

Explanation:

Given that

mass of the crate, m = 3 kg

angle of inclination, = 35°

coefficient of static friction, = 0.3

To solve this, we can assume that the minimum force is F Newton, then use the formula

mgsinA = coefficient of static friction * [F + mgcosA]

=>3 * 9.8 * sin35 = 0.3 * [F + 3 * 9.8 * cos35]

=> 29.4 * 0.5736 = 0.3 * [F + 29.4 * 0.8192]

=> 16.86 = 0.3 [F + 24.08]

=> 16.86 = 0.3F + 7.22

=> 16.86 - 7.22 = 0.3F

=> 0.3F = 9.64

=> F = 9.64/0.3

=> F = 32.13 N

Therefore, the Force that must be applied is 32.13 N

the average speed is 24.57 km/hr

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