A 32 kg block slides up a hill to a height of 6.4 m. If 266 J of thermal energy are generated, how fast was the block going at t
1 answer:
Answer:
Speed of the block at the bottom of the block is 11.91 m/s.
Explanation:
Energy can neither be created nor be destroyed but it can change its form from one to another such as kinetic energy to potential energy or vice a versa.
Applying conservation of energy to the given problem :
Kinetic Energy = Potential Energy + Thermal Energy
Here m is mass of the block, h is height, g is acceleration due to gravity and v is speed of the block.
Rearrange the above equation in terms of v ;
Substitute 32 kg for m, 6.4 m for h and 9.8 m/s² for g in the above equation.
<em>v = </em>11.91 m/s
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(a) Zero
When the ball reaches its highest point, the direction of motion of the ball reverses (from upward to downward). This means that the velocity is changing sign: this also means that at that moment, the velocity must be zero.
This can be also understood in terms of conservation of energy: when the ball is tossed up, initially it has kinetic energy
where m is the ball's mass and v is the initial speed. As it goes up, this kinetic energy is converted into potential energy, and when the ball reaches the highest point, all the kinetic energy has been converted into potential energy:
where g is the gravitational acceleration and h is the height of the ball at highest point. At that point, therefore, the potential energy is maximum, while the kinetic energy is zero, and so the velocity is also zero.
(b) 9.8 m/s upward
We can find the velocity of the ball 1 s before reaching its highest point by using the equation:
where
a = g = -9.8 m/s^2 is the acceleration due to gravity, which is negative since it points downward
v = 0 is the final velocity (at the highest point)
u is the initial velocity
t = 1 s is the time interval
Solving for u, we find
and the positive sign means it points upward.
(c) -9.8 m/s
The change in velocity during the 1-s interval is given by
where
v = 0 is the final velocity (at the highest point)
u = 9.8 m/s is the initial velocity
Substituting, we find
(d) 9.8 m/s downward
We can find the velocity of the ball 1 s after reaching its highest point by using again the equation:
where this time we have
a = g = -9.8 m/s^2 is the acceleration due to gravity, still negative
v is the final velocity (1 s after reaching the highest point)
u = 0 is the initial velocity (at the highest point)
t = 1 s is the time interval
Solving for v, we find
and the negative sign means it points downward.
(e) -9.8 m/s
The change in velocity during the 1-s interval is given by
where here we have
v = -9.8 m/s is the final velocity (1 s after reaching the highest point)
u = 0 is the initial velocity (at the highest point)
Substituting, we find
(f) -19.6 m/s
The change in velocity during the overall 2-s interval is given by
where in this case we have:
v = -9.8 m/s is the final velocity (1 s after reaching the highest point)
u = +9.8 m/s is the initial velocity (1 s before reaching the highest point)
Substituting, we find
(g) -9.8 m/s^2
There is always one force acting on the ball during the motion: the force of gravity, which is given by
where
m is the mass of the ball
g = -9.8 m/s^2 is the acceleration due to gravity
According to Newton's second law, the resultant of the forces acting on the body is equal to the product of mass and acceleration (a), so
which means that the acceleration is
and the negative sign means it points downward.