Answer:
The frictional force 6.446 N
The acceleration of the block a = 6.04
Explanation:
Mass of the block = 3.9 kg
°
= 0.22
(a). The frictional force is given by
3.9 × 9.81 ×
29.3 N
Therefore the frictional force
0.22 × 29.3
6.446 N
(b). Block acceleration is given by
F = 30 N
= 6.446 N
= 30 - 6.446
= 23.554 N
The net force acting on the block is given by
23.554 = 3.9 × a
a = 6.04
This is the acceleration of the block.
By using first and third equation of motion, the number of revolutions the tire makes during this motion is 43 rev.
ANGULAR MOTION
Since the car accelerate from rest, initial velocity will be zero.
Given that a car accelerates uniformly from rest and reaches a speed of 24.3 m/s in 9.1 s. The following are the given parameters
If the diameter of a tire is 80.2 cm, to find the number of revolutions the tire makes during this motion, we must first calculate the distance travelled by the car by using first and third equation of motion of the car.
First equation
V = U + at
Substitute all necessary parameters into the equation.
24.3 = 0 + 9.1a
a = 24.3/9.1
a = 2.67 m/
Third Equation of motion
Substitute all the necessary parameters
= 0 + 2 x 2.67 x S
590.49 = 5.34S
S = 590.49 / 5.34
S = 110.58 m.
Given that the diameter of a tire is 80.2 cm,
the radius (r) will be 80.2/2 = 40.1 cm
convert it to meter
r = 40.1/100 = 0.401 m
The Circumference of the tire = 2r
Circumference = 2 x 3.143 x 0.401
Circumference = 2.52 m
Assuming no slipping, number of revolutions = 110.58/2.52
Number of revolutions = 43.89 rev.
Number of revolutions = 43 rev.
Therefore, the number of revolutions the tire makes during this motion is 43 rev.
Learn more about circular motion here: brainly.com/question/6860269