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
<span>At the top of the waterfall, the water has potential energy. Once it goes over</span>
GPE= weight•height= 15 N• 0.22meter= 3.3 Joules
I hope this helps ~~Charlotte~~
Answer:
A
Explanation:
if the man doubles his force to 40 and the box was yet to move that means acceleration also doubled so your answer would be A
Answer:
The acceleration of the train is 5 m/s².
Explanation:
Given:
let the initial velocity of a train = 5 m/s and
final velocity of a train = 45 m/s
time taken = 8 s
To find:
acceleration: ?
Solution:
We define acceleration as change in velocity per unit time that is the difference between the final velocity and initial velocity divided by time.

On substituting the above values we get the required acceleration

Therefore,the acceleration of the train is 5 m/s².