Answer to Question #108278 in Mechanics | Relativity for lolipop

Question

[img]https://upload.cc/i1/2020/04/06/huqGD6.jpg[/img]


d8 is 4

Answer 1

Problem: A block weigth is M=2.4 kg. Spring force constant is k=2500N/m. Initial point A away from point O is $d_A=0.2m$. Coefficient of sliding friction is $\mu_k=0.3$ . Weigth of lump clay is $m=0.25 kg$ . Coefficient of static friction is $\mu_s=0.4$.

Solution (a): To find the speed of the block when it come to point O we can use the law of conservation mechanical energy with taking into account the work of the sliding friction force. The potential energy of spring at point A is $E_p=k\frac{d_A^2}{2}$ . The potential energy of the spring at point O is 0. The kinetic energy of block before lump clay fall is $E_k=\frac{MV_O^2}{2}$ . The work of the friction force is $W=-\mu_k\cdot Mg\cdot d_A$. The work of the friction force is always negative, because its direction is always opposite to the displacement of the body. Now we can write the law of conservation of energy

(1) $E_p+W=E_k$ Finding the block speed at point O.

(2) $V_O=\sqrt{\frac{k}{M}d_A^2-2\mu_k\cdot g\cdot d_A}=\sqrt{\frac{2500}{2.4}(0.2)^2-2\cdot 0.3\cdot9.8\cdot0.2}=\\=\sqrt{41.66-1.18}=6.36 m/s$

Solution (b): First of all, we should find the speed of movement of the block - drop glue system after an inelastic collision. To do this, we will have to use the law of conservation of momentum when moving in a horizontal direction. We believe that this interaction occurred so quickly that the influence of the friction force can be ignored. Impulse block before collision is $P_O=MV_O$. The momentum drop of glue in the horizontal direction before the collision is 0. The momentum of the block with glue after a collision is $P_1=(M+m)\cdot V_1$. The law of conservation of momentum is

$P_O=P_1$ and the velocity of system block-clay at point O is $V_1=\frac{M}{M+m}V_O=\frac{2.4}{2.4+0.25}\cdot6.36m/s=5.76m/s$. Now we can write the law of conservation of energy similarly to item (a). The kinetic energy of block with lump clay fall is $E_k=\frac{(M+m)V_1^2}{2}$ . The work of the friction force is $W=-\mu_k\cdot (M+m)g\cdot d_B=-0.3\cdot 2.65\cdot9.8\cdot d_B=-7.8Jm^{-1}d_B$ .

The potential energy of spring at point $d_B$ is $E_p=k\frac{d_B^2}{2}$. The law of conservation of energy is

(3) $E_k+W=E_p$ To find the block distance at point B we should solve the quadratic equation

(4) $kd_B^2+2\mu_k(M+m)g\cdot d_B -(M+m)V_1^2=0$

(5) $d_B=\frac{-\mu_k(M+m)g\pm\sqrt{(\mu_k(M+m)g)^2+k(M+m)V_1^2}}{k}=\frac{-7.8\pm\sqrt{60.7+2500\cdot 2.65\cdot 33.2 }}{2500}=0.18m$

A root with a negative sign in (5) should be thrown back; it does not correspond to the selected sign of displacement.

Solution (c): Find the force of spring at point $d_B$ . The force is $F_s=kd_B=2500\cdot 0.18=450N$. The maximum force of static friction is $F_f=\mu_s\cdot (M+m)\cdot g=0.4\cdot2.65\cdot 9.8=10.4N$. We see that $F_s>F_f$. This means that the block will not remain at rest at the extreme point B, but will begin to move in the opposite direction from point B to point O.

Answers:

(a) The speed of the block when it comes to O from A just before the clay sticks to it is 6.36m/s.

(b) Compression distance of the spring is 0.18m

(c) The system will begin to move in the opposite direction from point B to point O after the spring compressed.