3/28/2018 11:59 PM 25/100 3/27/2018 07:05 PM Gradebook Print Calculator Question

Question

3/28/2018 11:59 PM 25/100 3/27/2018 07:05 PM Gradebook Print Calculator Question 4 of 26 Incorrect Map Sapling Learning macmillan learning A 0.250-kg lump of clay is dropped from a height of 1.45 m onto the floor. It sticks to the floor and does not bounce What is the magnitude of the impulse J imparted to the clay by the floor during the impact? Number J1.33 kg m/s The force exerted by the floor on the clay is plotted as a function of time in the figure to the right. What must have been the maximum force Fx exerted by the floor on the clay? Number F532 5.5 ms Incorrect. Previous Give Up & View Solution # Try Again Next Exit Explanation

Explanation / Answer

Impulse is given by:

J = m*dV

J = m*(Vi - Vf)

Vf = 0, m = 0.25 kg

Using equation

Vi^2 = U^2 + 2*g*H

Vi = sqrt (2*9.81*1.45) = 5.33 m/sec

J = 0.25*5.33 = 1.33 kg-m/sec

Force is given by:

Fmax = J/dt

Fmax = 1.33/(5.5*10^-3)

Fmax = 241.82 N

Related Questions

Navigate

• Browse (All)
• Browse 3
• Subjects

---

---

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