Print Calculator-Periodic Table Question 4 of 14 Map Sapling Learning A 0.350-kg

Question

Print Calculator-Periodic Table Question 4 of 14 Map Sapling Learning A 0.350-kg lump of clay is dropped from a height of 1.75 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 kg m/s max 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 Fmax exerted by the floor on the clay? Number max 7.5 ms Previous Give Up & View Solution C Check Answer Next Exit Hint

Explanation / Answer

Here,

after falling by 1.75 m

speed of the lump = sqrt(2gh)

speed of the lump = sqrt(2 * 9.8 * 1.75)

speed of the lump = 5.86 m/s

Impulse imparted = m * v

Impulse imparted = 0.350 * 5.86

Impulse imparted = 2.05 Kg.m/s

-----------------------------------

area under the cruve = IMpulse

0.50 * 7.5 *10^-3 * Fmax = 2.05

Fmax = 547 N

the maximum force Fmax is 547 N

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