A 4 kg car with frictionless wheels is on a ramp that makes a 25° angle with the horizontal. The car is connected to a 1 kg mass
by a string that passes over a frictionless pulley.
What is the acceleration of the car?
What is the acceleration of the car?
1 answer:
Refer to the diagram shown below.
W₁ = (4 kg)*(9.8 m/s²) = 39.2 N
W₂ = (1 kg)*(9.8 m/s²) = 9.8 N
The normal reaction on the 4-kg mass is
N = (39.2 N)*cos(25°) = 35.5273 N
The force acting down the inclined plane due to the weight is
F = (39.2 N)*sin(25°) = 16.5666 N
The net force that accelerates the 4-kg mass at a m/²s down the plane is
F - W₂ = (4 kg)*(a m/s²)
4a = 16.5666 - 9.8
a = 1.6917 m/s²
Answer: 1.69 m/s² (nearest hundredth)
W₁ = (4 kg)*(9.8 m/s²) = 39.2 N
W₂ = (1 kg)*(9.8 m/s²) = 9.8 N
The normal reaction on the 4-kg mass is
N = (39.2 N)*cos(25°) = 35.5273 N
The force acting down the inclined plane due to the weight is
F = (39.2 N)*sin(25°) = 16.5666 N
The net force that accelerates the 4-kg mass at a m/²s down the plane is
F - W₂ = (4 kg)*(a m/s²)
4a = 16.5666 - 9.8
a = 1.6917 m/s²
Answer: 1.69 m/s² (nearest hundredth)
You might be interested in
Answer:
The ratio of their orbital speeds are 5:4.
Explanation:
Given that,
Mass of A = 5 m
Mass of B = 7 m
Radius of A = 4 r
Radius of B = 7 r
The orbital speed of satellite A,
......(I)
The orbital speed of satellite B,
......(I)
We need to calculate the ratio of their orbital speeds
Using equation (I) and (II)
Put the value into the formula
Hence, The ratio of their orbital speeds are 5:4.