Newton’s Second Law and the Work-Kinetic Energy Theorem October 13, 2010 Abstract This experiment utilizes an air track first as an inclined plane with the slider accelerating due to gravity and second as a level surface with the slider accelerating due to the pull of an attached free-falling object of known mass. In both cases, the Work performed is calculated based on formulas for mechanical work and for kinetic energy. The two results are compared. The first part yielded an average acceleration of 0.715 m/s2 (a 1.58% error) and the average result for the Work performed was 0.0204 N*m with only a 0.9% difference. The second part suffered critical errors due to improper data and the results are not significant or useful.
It can be expressed as a mathematical equation: or FORCE = MASS X ACCELERATION 3. “For every action there is an equal and opposite reaction.” This means that for every force there is a reaction force that is equal in size, but opposite in direction. That is to say that whenever an object pushes another object it gets pushed back in the opposite direction equally hard. The rocket's action is to
Since acceleration is a vector quantity it has magnitude and direction [3]. Uniformly accelerated movement (UAM) is movement that always has the same acceleration, meaning that it has a constant equal force. An object with a constant acceleration should not be confused with an object with a constant velocity; an object with a constant velocity is not accelerating. The perfect example of constant force is gravity (g Earth = 9.81 m/s2), which is the force of attraction between all masses in the universe, and in this cases specially the attraction of the earth's mass for bodies near its surface [4]. If an object is held motionless in a uniform gravitational field it will fall in a constant acceleration, this means that every second that the object falls its velocity will
But from here you may wonder what terminal velocity is? The answer is, terminal velocity is the velocity of an object when the sum of its drag force equals the downward force of weight that is acting on the object and hence meaning the object has no resulting acceleration and is moving at a constant velocity. However through research what I found is that you cannot 100% reach the terminal velocity of an object, but only x%(where x is any number between 0 and 100) of an objects terminal velocity as terminal velocity is asymptotic. Though this information is true, we still refer to an object that is falling with constant motion as reaching its terminal velocity and thus say that any object can and does reach terminal velocity. The way the viscosity of a liquid affects terminal velocity refers to the second statement that an object does reach terminal velocity.
Because ‘F’ and ‘x’ are directly proportional, a graph of ‘F’ vs ‘x’ is a line with slope ‘k’ A mass on a spring is a simple harmonic oscillator which is an object that oscillates the equilibrium point and experiences a restoring force proportional to the object’s displacement. The time it takes for a spring to complete an oscillation is called the period of oscillation, ‘T’. The period of oscillation of a simple harmonic oscillator that is described by Hooke’s Law is: T=2π√(m/k). This formula shows that as the mass, ‘m’, increases and the spring
The direction of acceleration is the same as the direction of the net force. The acceleration of the body is also directly proportional to the net force but inversely proportional to its mass. Newton defined momentum P as the product of mass and velocity. The change in momentum, symbolized by ∆P, is brought about by the impulse acting on the body, F_net ∆t=∆P As ∆t approaches zero, the instantaneous rate of change of momentum is, F_net=lim┬(∆t→0)〖∆P/∆t〗=dP/dt=(d(mv))/dt Since for most object, mass is constant, F_net=m dv/dt Newton’s second law of motion is mathematically expressed as F_net=ma From Newton’s second law T=m_1 a The hanging mass m_1 is also accelerating with the same acceleration due to the net force m_2 a on it. m_2 a=m_2 g-T T=m_2 g-m_2 a Equating the tensions m_1 a=m_2 g-m_2 a m_1 a+m_2 a=m_2 g (m_1+m_2 )a=m_2 g a=(m_2 g)/(m_1+m_2 ) The acceleration is the same acceleration described in the kinematics equation a=2s/t^2 For a body starting from rest, s is the distance traveled by the cart and t is the time of travel.
We used a vernier caliper to obtain the diameter of those two and therefore, the radius. When adding all the numbers together, we found that the true radius(r) of the orbit was 0.139 m. To find our tension, we needed to find out how much weight we needed to pull the object towards away from the spring and on the tip of the pointer as shown below. The tension needed to pull the mass on the tip of the pointer 1.05 kg. In theory the force of acceleration needed to pull the mass to same exact spot should equal the force of tension multiplied by the force due to gravity. Using Newton’s second law, F=ma, we know that the
In order to use the principles of fluid statics to analyze pressure in a system, it is helpful to make several assumptions. Fluids are assumed to be incompressible, in other words, they occupy a constant volume and maintain a constant density throughout the experiment. Pressure on a fluid at rest is assumed to be isotropic at every point, which is necessary to satisfy the zero sum force balance at a given point. In addition, pressure is assumed to be exerted normal to the contact surfaces at the boundaries of the system. Pressure in a given system is governed by the following equation: P = ρgh = γh (Eq.
Usually the experimenter adjusts the direction of the three forces, makes measurements of the amount of force in each direction, and determines the vector sum of three forces. Forces perpendicular to the plane of the force board are typically ignored in the analysis. In order to complete this lab we used a force table, accessories, level,standard weights. And weight hangers. In order to complete the first lab we had to level the table and connect the rings to the pulleys.
The velocity can be obtained by finding the slope of the graph of position as a function of time. The acceleration can be obtained by finding the slope of the graph of velocity as a function of time. The critical concepts are contained in the equations for motion with constant acceleration in one dimension, as follows: x=x0+vxot+1/2axt2 Equation 1 vx=vx0+axt Equation 2 In these equations, x is the position at time t andx0 is the position at time t=0 of the object; vxis the velocity of the object along the direction of motion, x, at time t, and is the velocity of the object along the direction of motion, x, at time t=0 ; and ax is the acceleration of the object along the direction of motion, x. Uniformly accelerated linear motion is all around us. Architects often consider the safety of the slides by simulating and calculating the acceleration of a child slides down.