What is the maximum value the string tension can have before the can slips? The coefficient of static friction between the can a
nd the ground is 0.76.
1 answer:
Answer:
T= 38.38 N
Explanation:
Here
mass of can = m = 3 kg
g= 9.8 m/sec2
angle θ = 40°
From figure we see the vertical and horizontal component of tension force T
If the can is to slip - then horizontal component of tension force should become equal to force of friction.
First we find force of friction
Fs= μ R
where
μ = 0.76
R = weight of can = mg = 3 × 9.8 = 29.4 N
Now horizontal component of tension
Tx= T cos 40 = T× 0.7660 N
==>T× 0.7660 = 29.4
==> T= 38.38 N
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Answer: A) Deceleration of the car is -6.6667 m/s² while it came to stop.
B) The total distance the car travels is 200 meter during the 10 s period.
Explanation:
Given Data
Initial velocity of the car () = 20.0 m/s
Final velocity of the car () = 0 m/s
Time (in motion) =7.00 s
Time (in rest) =3 s
To find - A) car's deceleration while it came to a stop
B) the total distance the car travels in 10 s
A) The formula to find the deceleration is
Deceleration = (( final velocity - initial velocity ) ÷ Time) (m/s²)
Deceleration = (() - (
)) ÷ time (m/s²)
Deceleration = ( 0.0 - 20 ) ÷ 3 (m/s²)
Deceleration = (- 20) ÷ 3 (m/s²)
Deceleration = - 6.6667 m/s²
(NOTE : Deceleration is the opposite of acceleration so the final result must have the negative sign)
The car's deceleration is - 6.6667 m/s² while it came to a stop
B) The formula to find the distance traveled by the car is
Distance traveled by the car is equals to the product of the speed and time
Distance = Speed × Time (meter)
Distance = 20.0 × 10
Distance = 200 meters
The total distance the car travels during the period of 10 s is 200 meters
Explanation:
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