Let the angle be Θ (theta)
Let the mass of the crate be m.
a) When the crate just begins to slip. At that moment the net force will be equal to zero and the static friction will be at the maximum vale.
Normal force (N) = mg CosΘ
μ (coefficient of static friction) = 0.29
Static friction = μN = μmg CosΘ
Now, along the ramp, the equation of net force will be:
mg SinΘ - μmg CosΘ = 0
mg SinΘ = μmg CosΘ
tan Θ = μ
tan Θ = 0.29
Θ = 16.17°
b) Let the acceleration be a.
Coefficient of kinetic friction = μ = 0.26
Now, the equation of net force will be:
mg sinΘ - μ mg CosΘ = ma
a = g SinΘ - μg CosΘ
Plugging the values
a = 9.8 × 0.278 - 0.26 × 9.8 × 0.96
a = 2.7244 - 2.44608
a = 0.278 m/s^2
Hence, the acceleration is 0.278 m/s^2
Answer:
kinetic energy + potential energy
I think its suicidal ideation......
I think
Potential Energy= 24m * 14kg * 9.8N/kg = 3292.8J
Well it seems that you did not give answer choices, but that its fine since we can use newtons law of universal gravitational, Fg = GM1M2/r^2. So G is the gravitational constant, which is 6.67*10^-11, we can plug in 6*1024 for M1, and 7*1022 for M2, and 3.8*108 for r. Which then we get 1.74 * 10^8 N as the force of attraction between the Earth and the moon.
