Please Show full steps problem 1 : A ball of mass m = 0.2 kg has an x-position a

Question

Please Show full steps

problem 1 : A ball of mass m = 0.2 kg has an x-position as a function of time of

x=(10m)-(2m/s^2)t^2+(0.5m/s^4)t^4

A) Find the velocity of the ball as a function of time.

B) What is the initial velocity of the ball?

C) Find the velocity of the ball at time t = 2s. D) Find the acceleration of the ball as a function of time.

D) Find the acceleration of the ball as a function of time

E) Find the net force on the ball as a function of time.

F) Find the net force on the ball at time t = 2s.

Problem 2: What must be the initial speed of a ball launched at an angle of 30 degree ? from ground level if it is to be at a height of 10 m when it is a horizontal distance of 50 m from the launch point?

Problem 3: Consider a block of mass 4 kg on a horizontal surface. A pulling force is directed at an angle of 30 degree ? above the horizontal:.

A) If the pulling force has a size of 15 N and the surface is frictionless, find the acceleration of the block.

B) If the pulling force has a size of 100 N and the surface is frictionless, finnd the acceleration of the block.

Explanation / Answer

a) v = dx/dt = - 4 t + 2 t^3

b) v(0) = 0

C) v(2) = -8 + 2*8 = 8

D) a = dv/dt = -4 + 6 t^2

E) F = ma = 0.2*(-4 + 6 t^2) = -0.8 + 1.2 t^2

F) F(2) = -0.8 + 1.2*4=4

2)

x direction

x = v cos(30) t

t = 50/(v cos(30))

y direction

10 = v*sin(30)*t - 0.5*9.81 * t^2

10 = v*sin(30)*50/(v cos(30)) - 0.5*9.81 * (50/(v cos(30)))^2

v=29.44 m/s

3) sum forces in the x

15*cos(30) = 4*a

a = 15*cos(30 degrees)/4=3.25

B) 100*cos(30 degrees)/4=21.65 m/s^2

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