A ball of mass 0.1 kg is attached to a rubber band, which we can model as a spring of stiffness 8 N/m. With one end of the rubber band fixed, you throw the ball vertically upward with an initial speed of 14 m/s. At the moment the ball leaves your hand, the rubber band is already stretched 0.1 m from its unstretched position. What is the maximum height the ball will rise (relative to where it was thrown)?
initial KE+initialPEspring=finalKE+finalPEspring+finalgpe
1/2 m 14^2+1/2 k .1^2=0+1/2 k h^2+mgh
solve for h.
A 0.4 RUBBER BALL FALLS TO THE GROUND AT 8.0 M/S. IF IT BOUNCES BACK UPWARDS AT THE SAME SPEED,WHAT IS THE IMPULSE ACTING ON THE BALL?
To find the maximum height the ball will rise, we need to use the principle of conservation of mechanical energy. The mechanical energy of the ball is conserved throughout its motion.
First, let's find the spring potential energy when the ball is released from your hand. The spring potential energy can be calculated using the formula:
PE_spring = (1/2)kx^2
where k is the stiffness of the rubber band (8 N/m) and x is the displacement of the rubber band from its unstretched position (0.1 m).
PE_spring = (1/2)(8 N/m)(0.1 m)^2
= 0.04 J
Next, let's find the initial kinetic energy of the ball when it leaves your hand. The initial kinetic energy can be calculated using the formula:
KE_initial = (1/2)mv^2
where m is the mass of the ball (0.1 kg) and v is the initial velocity of the ball (14 m/s).
KE_initial = (1/2)(0.1 kg)(14 m/s)^2
= 9.8 J
According to the conservation of mechanical energy, the sum of the initial kinetic energy and the initial potential energy (spring potential energy) is equal to the sum of the final kinetic energy and the final potential energy (which is equal to zero at the maximum height).
KE_initial + PE_spring = KE_final + PE_gravitational
Since the ball reaches its maximum height, its final velocity is 0 m/s, and therefore its final kinetic energy is zero.
KE_final = 0 J
Using the formula for gravitational potential energy:
PE_gravitational = mgh
where m is the mass of the ball (0.1 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the maximum height.
0.04 J = 0 J + (0.1 kg)(9.8 m/s^2)h
0.04 J = 0.98 Jh
h = 0.04 J / 0.98 J
= 0.041 m
Therefore, the maximum height the ball will rise (relative to where it was thrown) is approximately 0.041 meters.