Find the percentage of the total work lost to friction if 28.7 J of work is put into pushing a block up a ramp resulting in 14.2
1 answer:
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Answer:
Shortest time = 3.47 seconds
Explanation:
We know that Force of gravity of an object is given by;
F_g = mg
Also, Force due to kinetic friction is given as;
F_friction = μ_k•F_n
Force due to static friction is given as;
F_static = μ_s•F_n
Where ;
μ_k = coefficient of kinetic friction
μ_s = coefficient of static friction
F_n = Normal Force
Let's solve for the static crate.
From the free body diagram i attached, the resultant force in the vertical y direction will be;
N - mg = 0
Thus, N = mg
WherN is the
From earlier, we know that;
F_static = μ_s•F_n
Thus,
F_static = μ_s•mg
Now, in the horizontal x direction,
F_static = ma
Thus,
μ_s•mg = ma
m will cancel out, and;
a = μ_s•g
a = 0.3 x 9.81 = 2.943 m/s²
Not, to get the final the final speed of the tool box, we'll use;
V_f = V_i + a•Δt
Where;
Δt is shortest time.
V_f = final velocity
V_i = initial velocity
Thus, making Δt the subject, we have;
Δt = (V_f - V_i)/a = (10.2 - 0)/2.943 = 3.47 s
The magnitude of the acceleration of the ball while coming to rest is 477.43 m/s²
The direction of the acceleration of the ball is downwards
The given parameters
initial velocity of the ball, u = 0
height above the ground, h = 2.2 m
time of motion of the ball, t = 96 ms = 0.096 s
The magnitude of the acceleration of the ball while coming to rest is calculated as;
let the downwards direction of the acceleration be positive
The direction of the acceleration of the ball is downwards
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