The force required to push the impala down on ground is 1,500 N.The ratio of force to weight is 3.4.
Answer:
Explanation:
The mass of the impala is given as 45 kg, the height or displacement of the impala is 2.5 m and it took 0.21 s. As the impala was moved straight up, so the acceleration due to gravity will be working as acceleration but in opposite direction.
As the initial velocity u is given as 0 and the final velocity we have to find
So v = 7 m/s.
As the final velocity is 7 m/s and the time is 0.21 s, then acceleration is rate of change of velocity with time.
As per Newton's second law of motion, the force acting on the impala will be equal to the product of mass and acceleration of the impala
F = Mass * Acceleration = 45 * 33.333=1,500 N.
So the force required to push the impala down on ground is 1,500 N.
And the weight is measured as product of acceleration due to gravity with mass.
Weight = 45 * 9.8 = 441.
So the ratio of force to weight is 1500/441 = 3.4
Thus, the ratio of force to weight is 3.4.
Answer:
The speed of washer will be 32.144 m/s
Explanation:
Given that
time t= 3.28 s
Acceleration due to gravity
Here acceleration is constant so we cam apply the motion equation
we know that
v= u + at
At initial condition u =0 m/s
v= u + at
v= 0 + 9.8 x 3.28
v= 32.144 m/s
So the speed of washer will be 32.144 m/s
Answer:
The correct option is;
d) Parallel to the plane
Explanation:
The forces acting on the box of mass, m are;
Weight of the box acting an angle, θ, equal to the inclination of the plane to the perpendicular of the plane
Weight of the box acting along the plane = m×g×sin(θ)
The force of friction along the plane = μ×m×g×cos(θ)
The total force acting downward along the plane , = m×g×sin(θ) + μ×m×g×cos(θ)
The Force needed to pull the box up along the plane F = The total force acting downward along the plane
F = m×g×sin(θ) + μ×m×g×cos(θ) = m×g×(sin(θ) + μ×cos(θ))
When the force, Fₐ is applied vertically, the force acting along the plane = Fₐ×cos(θ)
When the force is applied perpendicular, the force acting along the plane = Fₐ×sin(θ)
When the force is applied horizontally, the force acting along the plane = Fₐ×cos(θ)
When the force is applied parallel to the plane, the force acting along the plane = Fₐ
Therefore, since Fₐ > Fₐ×cos(θ) and Fₐ > Fₐ×sin(θ), for acute angles, we have that the least force is required when the force is acting parallel to the plane.