Am sending the same question coz i can't find the first one on the message boards. So plz help me.
1. 'velocity is a vector quantity, and the kinetic energy of a body is equal to half its mass times the square of its velocity, so kinetic energy must also be a vector.' Comment on the correctness of this argument.
2. A trolley of mass 0.80kg runs freely, without accelerating down an inclined plane when the plane makes 5 degrees with the horizontal. Finad the force parallel to the plane resisting motion.
The answer for this question is 0.68N. But by calculation I can only find the driving force of the trolley which is 8cos5. It is the vetical component of the weight which gives 0.68N and it is not parallel to the plane. So how can I calculate the frictional force??
Plz help!!!
Kinetic energy is not a vector. It is proportional to the square of velocity, regardless of direction.
Since the trolley is not accelerating, the friction force exactly balances the component of the gravity force down the ramp, which is M g sin 5 = 0.68 N
A+b
1. The argument that kinetic energy must be a vector because velocity is a vector is incorrect. Kinetic energy is a scalar quantity, not a vector. While velocity is a vector because it has both magnitude and direction, kinetic energy only depends on the magnitude of velocity, not its direction. The formula for kinetic energy, which is half the mass times the square of the velocity, only involves scalar quantities (mass and velocity). Therefore, kinetic energy is not affected by the direction of velocity and remains a scalar.
2. To calculate the force parallel to the plane resisting motion, we need to consider the forces acting on the trolley. The two main forces in this case are the component of the weight of the trolley down the inclined plane and the friction force opposing motion.
The component of the weight down the plane can be calculated using the formula mg sin θ, where m is the mass of the trolley and θ is the angle the plane makes with the horizontal. In this case, m = 0.80 kg and θ = 5 degrees. So the component of the weight down the plane is:
Force_down_plane = m * g * sin θ = 0.80 kg * 9.8 m/s^2 * sin 5° ≈ 0.68 N
Since the trolley is not accelerating, the friction force opposing motion must be equal in magnitude and opposite in direction to the force down the plane. So the force parallel to the plane resisting motion is also approximately 0.68 N.
The driving force of the trolley, which is the vertical component of the weight, is not relevant to calculating the frictional force.
Therefore, the answer for the force parallel to the plane resisting motion is indeed 0.68 N.