Please show me how to work the following problems.
1) A cyclist whose mass is 48 kg turns a corner with a radius of 30 m at a speed of 25 m/s.
A) What is the magnitude of the cyclist's acceleration?
B) What is the magnitude of the net force acting on the cyclist?
2) A rock is thrown off a tall cliff with a vertical speed of 53 m/s upward and a horizontal speed of 30 m/s.
A) What will these speeds Vx and Xy, be 4 second later?
B) What is the x and y displacements of the rock be 4 seconds later?
3) A bowling ball rolls off the edge of a giant' table at 18 m/s. If it takes 6 s for the ball to hit the ground.
A) How far does it fall?
Given: Vx = 30 m/s., Vy = 53 m/s.
2A. Vx = 30 m/s.
Vy = 53 + g*t = 53 - 9.8*4 = 13.8 m/s.
2B. Dx = Vx*t = 30 * 4 = 120 m.
Dy = 13.8 * 4 = 55.2 m.
3A. h = 0.5g*t^2 = 4.9*6^2 = 176.4 m.
Sure! I can help you with these problems. Let's solve them step by step.
Problem 1:
A) To find the magnitude of the cyclist's acceleration, we can use the formula: acceleration = v^2 / r, where v is the velocity and r is the radius of the circular path.
Given:
Mass of the cyclist (m) = 48 kg
Radius of the corner (r) = 30 m
Speed of the cyclist (v) = 25 m/s
Using the formula, the acceleration is:
acceleration = (25 m/s)^2 / 30 m = 20.83 m/s^2
Therefore, the magnitude of the cyclist's acceleration is approximately 20.83 m/s^2.
B) To find the magnitude of the net force acting on the cyclist, we can use the formula: force = mass * acceleration.
Given:
Mass of the cyclist (m) = 48 kg
Acceleration (a) = 20.83 m/s^2 (as calculated in part A)
Using the formula, the net force is:
force = 48 kg * 20.83 m/s^2 = 999.84 N
Therefore, the magnitude of the net force acting on the cyclist is approximately 999.84 N.
Problem 2:
A) To find the speeds Vx (horizontal) and Vy (vertical) after 4 seconds, we can use the equations of motion.
Given:
Initial vertical speed (Vyi) = 53 m/s (upward)
Initial horizontal speed (Vxi) = 30 m/s
Time (t) = 4 seconds
Acceleration due to gravity (g) = 9.8 m/s^2 (assuming no air resistance)
Vertical speed after 4 seconds:
Vy = Vyi + g * t = 53 m/s + (9.8 m/s^2 * 4 s) = 53 m/s + 39.2 m/s = 92.2 m/s (upward)
Horizontal speed remains constant, so Vx remains the same:
Vx = Vxi = 30 m/s
Therefore, the speeds Vx and Vy after 4 seconds are 30 m/s (horizontal) and 92.2 m/s (upward), respectively.
B) To find the x and y displacements (horizontal and vertical distances) of the rock after 4 seconds, we can again use the equations of motion.
Given:
Initial vertical speed (Vyi) = 53 m/s (upward)
Initial horizontal speed (Vxi) = 30 m/s
Time (t) = 4 seconds
Acceleration due to gravity (g) = 9.8 m/s^2 (assuming no air resistance)
Vertical displacement after 4 seconds:
Sy = Vyi * t + 0.5 * g * t^2 = 53 m/s * 4 s + (0.5 * 9.8 m/s^2 * (4 s)^2) = 212 m + 78.4 m = 290.4 m (upward)
Horizontal displacement after 4 seconds:
Sx = Vxi * t = 30 m/s * 4 s = 120 m
Therefore, the x and y displacements of the rock after 4 seconds are 120 m (horizontal) and 290.4 m (upward), respectively.
Problem 3:
To find how far the bowling ball falls, we can use the equation: distance = initial velocity * time + 0.5 * acceleration * time^2.
Given:
Initial velocity (Vi) = 18 m/s
Time (t) = 6 seconds
Acceleration due to gravity (g) = 9.8 m/s^2 (assuming no air resistance)
Using the formula, the distance is:
distance = Vi * t + 0.5 * g * t^2 = 18 m/s * 6 s + (0.5 * 9.8 m/s^2 * (6 s)^2) = 108 m + 176.4 m = 284.4 m
Therefore, the bowling ball falls approximately 284.4 meters.
I hope this helps! Let me know if you have any further questions.