Physics Lab Report Essay

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In the chapter Kinematics of Particles, we studied about various relations among time, position, velocity and acceleration of a particle. When we have a relation between any two parameters, we can calculate remaining parameters. In Kinetics of Particles, we applied Newton’s second law of motion ƩF = ma. The fundamental concept is when external forces acting on a particle are balanced (zero resultant), the particle will be in equilibrium. But when external forces acting on a particle are not balanced, the particle will move in the direction of the resultant with magnitude ma. We use Newton’s second law to calculate the acceleration of a particle. Once the acceleration of a particle is known, we can apply equations of kinematics to obtain the …show more content…
2.3

Work by a spring force and gravity
Work by a spring force
Consider a liner spring of spring constant k and undeformed length L. An external force F acts on the spring, which stretches the spring from position 1 to 2 (Fig. 2.4). (a)

(b)

Fig. 2.4
The deformation of the spring at position 1 and 2 are x1 and x2 respectively. When the spring is stretched, the spring force (Fs) tries to restore the spring to its original position. So the spring force Fs acts opposite to the displacement. We also know that force developed in a liner spring is directly proportional to the deformation (we only consider the case of liner spring). Since the displacement of the spring is opposite to the force Fs, work done by the spring force is negative. So we can write dU= -F_s.dx
Integrating from position 1 to 2
∫_1^2▒dU=-∫_(x_1)^(x_2)▒〖F_s dx〗
We know that for a liner spring Fs = kx. Now we have
U_(1→2)=-∫_(x_1)^(x_2)▒kxdx=-k/2(x_2^2-x_1^2)
U_(1→2)=k/2(x_1^2-x_2^2) Eq. 2.4

Where, k = Spring constant (N/m) x1 = deformation of the spring measured from its undeformed length at initial position x2 = deformation of the spring measured from its actual length at final
…show more content…
The tangential component of the force is responsible for moving the particle along the path. From Newton’s second law of motion for normal and tangential axes, we have
F_t=ma_t
From the kinematic relation, we know that at = vdv/dx, so we can write the above equation as
Fig. 2.7

F_t=mv dv/dx
F_t dx=mvdv
We already know that the first term in the above equation is work due to a force and it is a scalar term. So we write the equation in scalar form. If the particle has initial position x1, initial velocity v1, final position x2 and final velocity v2, we have
∫_(x_1)^(x_2)▒〖F_t dx〗=∫_(v_1)^(v_2)▒mvdv
∫_(x_1)^(x_2)▒〖F_t dx〗=(mv_2^2)/2-(mv_1^2)/2
The first term in the above equation is the work done by a force and the second and the third terms are kinetic energies of a particle. Since the work due to a force is scalar quantity, kinetic energy of a particle is also a scalar quantity. We write above equation as
U_(1→2)=T_2-T_1
T_1+U_(1→2)=T_2 Eq. 2.10

Eq. 2.10 is the mathematical representation of the principle of work and energy for a particle. It states that initial kinetic energy of a particle plus total work done on the particle is equal to the final kinetic energy of the

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