Answer:
1.5 u
Explanation:
The range equation is:
R = u² sin(2θ) / g
When u = v, R = 2.25 R.
2.25 R = v² sin(2θ) / g
2.25 u² sin(2θ) / g = v² sin(2θ) / g
2.25 u² = v²
1.5 u = v
Answer:
Explanation:
Let the bigger crate be in touch with the ground which is friction less. In the first case both m₁ and m₂ will move with common acceleration because m₁ is not sliding over m₂.
1 ) Common acceleration a = force / total mass
= 234 / ( 25 +91 )
= 2.017 m s⁻².
2 ) Force on m₁ accelerating it , which is nothing but friction force on it by m₂
= mass x acceleration
= 25 x 2.017
= 50.425 N
The same force will be applied by m₁ on m₂ as friction force which will act in opposite direction.
3 ) Maximum friction force that is possible between m₁ and m₂
= μ_s m₁g
= .79 x 25 x 9.8
= 193.55 N
Acceleration of m₁
= 193 .55 / 25
= 7.742 m s⁻²
This is the common acceleration in case of maximum tension required
So tension in rope
= ( 25 +91 ) x 7.742
= 898 N
4 ) In case of upper crate sliding on m₂ , maximum friction force on m₁
= μ_k m₁g
= .62 x 25 x 9.8
= 151.9 N
Acceleration of m₁
= 151.9 / 25
= 6.076 m s⁻².
Answer:
Magnitude of the Frictional force = (mv₀²)/2x₁
Explanation:
For the frictional force to stop the box, it has to produce the deceleration of the box; thereby being the opposing force to the box's motion.
According to Newton's first law of motion
Frictional force = (mass of the box) × (deceleration experienced by the box)
Let the mass of the box be m
Then,
Frictional force = ma
Then we can obtain the deceleration using the equations of motion
v² = u² + 2ax
u = Initial velocity = v₀ m/s
v = Final velocity = 0 m/s (since the box comes to rest at the end)
x = horizontal distance covered = (x₁ - x₀) = x₁ (since x₀ = 0)
a = ?
v² = u² + 2ax
0 = (v₀)² + 2ax₁
2ax₁ = - v₀²
a = - (v₀²)/(2x₁) (minus sign, because it's a deceleration)
Magnitude of the Frictional force = ma = (mv₀²)/2x₁
Hi there!
On a level road:
∑F = Ff (Force due to friction)
The net force is the centripetal force, so:
mv²/r = Ff
Rewrite the force due to friction:
mv²/r = μmg
Cancel out the mass:
v²/r = μg
Solve for v:
v = √rμg
v = √(25)(9.81)(0.8) = 14.01 m/s
