Answer:
The box will be moving at 0.45m/s. The solution to this problem requires the knowledge and application of newtons second law of motion and the knowledge of linear motion. The vertical component of the force Fp acts vertically upwards against the directio of motion. This causes a constant upward force of 23sin45° to act on the box. Fhe frictional force of 13N also acts vertically upwards and so two forces act upwards against rhe force of gravity resulting un a net force of 0.7N acting kn the box. This corresponds to an acceleration of 0.225m/s². So in w.0s after i start to push v = 0.45m/s.
Explanation:
Answers are:
(1) KE = 1 kg m^2/s^2
(2) KE = 2 kg m^2/s^2
(3) KE = 3 kg m^2/s^2
(4) KE = 4 kg m^2/s^2
Explanation:
(1) Given mass = 0.125 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.125 * (16)
KE = 1 kg m^2/s^2
(2) Given mass = 0.250 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.250 * (16)
KE = 2 kg m^2/s^2
(3) Given mass = 0.375 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.375 * (16)
KE = 3 kg m^2/s^2
(4) Given mass = 0.500 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.5 * (16)
KE = 4 kg m^2/s^2
Answer:
0.0675 seconds
Explanation:
From the question,
We apply newton's second law of motion
F = m(v-u)/t.................... Equation 1
Where F = force exert by the brake, v = final speed, u = initial speed m = mass of the bicycle, t = time.
make t the subject of the equation
t = m(v-u)/F................... Equation 2
Given: m = 180 kg, u = 6.0 m/s, v = 0 m/s (comes to stop), F = -1600 N ( agianst the dirction of motion)
Substitute these value into equation 2
t = 180(0-6.0)/-1600
t = -1080/-1600
t = 0.0675 seconds.
The roughness of the surface, the mass of the object, and the area of contact.