- #1

jisbon

- 476

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- Homework Statement
- Block of mass m = 0.5kg sits on top of large plate with mass M = 1.5kg. Contact surface of plate and ground is friction less and coefficient of static and kinetic friction is 0.9 and 0.5 respectively. System is originally at rest and external force F(t) = 2t N (where t is time in seconds) is used to pull the block. Until block starts to slide, after which F is maintained at value in part (a). Assume plate is long enough such block doesn't fall off plate.

(a) Calculate external force F at moment before block starts to slide on the plate.

(b) Calculate how far plate will travel relative to ground 1s after block slips.

(c) Calculate how far block travel relative to the ground 1s after it slips.

- Relevant Equations
- s=ut+0.5at^2

F=ma

Question as follows:

Force needed to make block slide = static friction between block and plate.

So

To find distance , use

s=ut+0.5at2 where u=0, so s=0.2at2s=0.2at2

To find a, F=ma

Since plate only experience kinetic friction(?),

a=7.35/2a=7.35/2 (including block on plate) = 3.6753.675

s=0.5(3.675)=1.8375s=0.5(3.675)=1.8375m to the left?

Do help explain if it's wrong. Thanks

a=8.82−2.45/0.5=12.74a=8.82−2.45/0.5=12.74

s=0.5(12.74)(1)2=6.37ms=0.5(12.74)(1)2=6.37m to the right?

Help will really be appreciated. Thanks

My answer/explanation:Block of mass m = 0.5kg sits on top of large plate with mass M = 1.5kg. Contact surface of plate and ground is friction less and coefficient of static and kinetic friction is 0.9 and 0.5 respectively. System is originally at rest and external force F(t) = 2t N (where t is time in seconds) is used to pull the block. Until block starts to slide, after which F is maintained at value in part (a). Assume plate is long enough such block doesn't fall off plate.

(a) Calculate external force F at moment before block starts to slide on the plate.

Force needed to make block slide = static friction between block and plate.

So

**friction = normal of block * static friction**= (0.5∗9.8)∗0.9=4.41N(0.5∗9.8)∗0.9=4.41NMy thoughts/wrong answer probably?: after block slips, plate experience(b) Calculate how far plate will travel relative to ground 1s after block slips.

**kinetic friction**= (1.5∗9.8)∗0.5=7.35N(1.5∗9.8)∗0.5=7.35NTo find distance , use

s=ut+0.5at2 where u=0, so s=0.2at2s=0.2at2

To find a, F=ma

Since plate only experience kinetic friction(?),

a=7.35/2a=7.35/2 (including block on plate) = 3.6753.675

s=0.5(3.675)=1.8375s=0.5(3.675)=1.8375m to the left?

Do help explain if it's wrong. Thanks

Using similar concepts as above, After 1s,(c) Calculate how far block travel relative to the ground 1s after it slips.

**F= 2t N**= 2(1)(4.41)=8.822(1)(4.41)=8.82**Friction**= 0.5∗0.5∗9.8=2.450.5∗0.5∗9.8=2.45a=8.82−2.45/0.5=12.74a=8.82−2.45/0.5=12.74

s=0.5(12.74)(1)2=6.37ms=0.5(12.74)(1)2=6.37m to the right?

Help will really be appreciated. Thanks