) A 1000-kg car is moving at 30 m/s around a horizontal unbanked curve whose diameter is 0.20 km. What is the magnitude of the f
1 answer:
The frictional force required is 9000 N
Explanation:
In order to keep the car in the turn in circular motion without sliding, the frictional force must provide the centripetal force necessary for the circular motion.
Therefore, we can write:
where the term on the left is the frictional force while the term on the right is the centripetal force, and where:
m is the mass of the car
v is its speed
r is the radius of the curve
For the car in this turn, we have
m = 1000 kg
v = 30 m/s
(since the diameter is 0.20 km, the radius is half that value)
And substituting, we find
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= velocity of migrating robin relative to air = 12 j m/s
(where "j" is unit vector in Y-direction)
= velocity of air relative to ground = 6.3 i m/s
(where "i" is unit vector in X-direction)
= velocity of migrating robin relative to ground = ?
using the equation
=
+
= 12 j + 6.3 i
= 6.3 i + 12 j
magnitude : sqrt((6.3)² + (12)²) = 13.6 m/s
direction : tan⁻¹(12/6.3) = 62.3 deg north of east
In order to determine the required force to stop the car, proceed as follow:
Calculate the deceleration of the car, by using the following formula:
where,
v: final speed = 0m/s (the car stops)
vo: initial speed = 36m/s
x: distance traveled = 980m
a: deceleration of the car= ?
Solve the equation above for a, replace the values of the other parameters and simplify:
Next, consider that the formula for the force is:
where,
m: mass of the car = 820 kg
a: deceleration of the car = 0.66m/s^2
Replace the previous values and simplify:
Hence, the required force to stop the car is 542.20N