# Physics Lab Report Essay

*…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