A crate that weighs 20N starts from rest and slides down arough 5.0m long ramp,

Question

A crate that weighs 20N starts from rest and slides down arough 5.0m long ramp, inclined at 25 degrees with thehorizontal. During this slide 20J of energy is lost due tofriction. a. What is the kinetic friction force on the crate? b. What is the speed of the crate at the bottom of theramp? A crate that weighs 20N starts from rest and slides down arough 5.0m long ramp, inclined at 25 degrees with thehorizontal. During this slide 20J of energy is lost due tofriction. a. What is the kinetic friction force on the crate? b. What is the speed of the crate at the bottom of theramp?

Explanation / Answer

       
   a.   Work done againstfriction   Wf   =   Forceof friction (Ff) * displacement (d)          20J   =    * W * cos *d       =   coefficientof kineticfriction,   W   =   Weight   =   20N,      =   Angleofincline   =   250,   d   =   5.0m    20   =   * 20 * cos 250 * 5.0       =   1 /4.53   =   0.221    b.   Netforce   F   =   W *sin   -   Ff   =   W* sin    -    *W * cos    =   20 * sin250   -   0.221 * 20 * cos250    F   =   4.45   N    Mass   m   =   W/ g   =   20 /9.8   =   2.04   kg    Acceleration   a   =   F/ m   =   4.45 /2.04   =   2.18   m/s    According to third equation    vf2   =   vi2   +   2* a * d    initialvelocity   vi   =   0    vf2   =   02   +   2* 2.14 * 5.0   =   21.4    final velocity   vf   =   21.4   =   4.63   m/s
   a.   Work done againstfriction   Wf   =   Forceof friction (Ff) * displacement (d)          20J   =    * W * cos *d       =   coefficientof kineticfriction,   W   =   Weight   =   20N,      =   Angleofincline   =   250,   d   =   5.0m    20   =   * 20 * cos 250 * 5.0       =   1 /4.53   =   0.221    b.   Netforce   F   =   W *sin   -   Ff   =   W* sin    -    *W * cos    =   20 * sin250   -   0.221 * 20 * cos250    F   =   4.45   N    Mass   m   =   W/ g   =   20 /9.8   =   2.04   kg    Acceleration   a   =   F/ m   =   4.45 /2.04   =   2.18   m/s    According to third equation    vf2   =   vi2   +   2* a * d    initialvelocity   vi   =   0    vf2   =   02   +   2* 2.14 * 5.0   =   21.4    final velocity   vf   =   21.4   =   4.63   m/s    F   =   4.45   N    Mass   m   =   W/ g   =   20 /9.8   =   2.04   kg    Acceleration   a   =   F/ m   =   4.45 /2.04   =   2.18   m/s    According to third equation    vf2   =   vi2   +   2* a * d    initialvelocity   vi   =   0    vf2   =   02   +   2* 2.14 * 5.0   =   21.4    final velocity   vf   =   21.4   =   4.63   m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.